MATH 334 - DIFFERENTIAL
EQUATIONS - Cauchy-Euler equations
Question Bank - Set 4
Liberty University
Question 1
Question
Solve the Cauchy-Euler equation (x2−4)y′′−3xy′+3y= 0 using the substitution
y=xr.
Solution
Step 1: Substitute y=xrinto the Cauchy-Euler equation (x2−4)y′′−3xy′+3y=
0 and find y′and y′′.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the equation.
(x2−4)r(r−1)xr−2−3xrxr−1+ 3xr= 0
Step 3: Simplify the equation to get rid of the xterms.
r(r−1)xr−3rxr+ 3xr= 0 =⇒r2−4r+ 3 = 0
Step 4: Solve the characteristic equation r2−4r+ 3 = 0 to find the roots r1
and r2.
(r−1)(r−3) = 0 =⇒r1= 1, r2= 3
Step 5: Write down the general solution to the Cauchy-Euler equation using
the roots r1= 1 and r2= 3.
y(x) = c1x+c2x3where c1, c2are arbitrary constants
Therefore, the solution to the Cauchy-Euler equation (x2−4)y′′−3xy′+3y=
0 is y(x) = c1x+c2x3.
Question 2
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
To solve the given Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we will assume
a solution of the form y=xrand then find the values of r.
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equa-
tion.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
Step 4: Solve the resulting quadratic equation for r.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 5: Form the general solution. Since r= 2, the solution to the
Cauchy-Euler equation is y=c1x2+c2x2ln x, where c1and c2are arbitrary
constants.
Question 3
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
2
Solution
Step 1: Assume a solution of the form y=xrwhere ris a constant to be
determined. Then, calculate the first and second derivatives of y.
Let y=xr
First derivative: y′=rxr−1
Second derivative: y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation, and simplify.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out an xrfrom the equation.
xr(r(r−1) −3r+ 4) = 0
Step 4: Set the expression inside the parentheses equal to zero to find the
characteristic equation.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 5: Find the roots of the characteristic equation.
r= 2
Step 6: Form the general solution using the roots found in Step 5.
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x), where c1and c2are arbitrary constants.
Question 4
Question
Solve the Cauchy-Euler equation x2y′′ −7xy′+ 12y= 0.
3
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Calculate the first and second derivatives of y.
y′=rxr−1, y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the original differential equation.
x2(r(r−1)xr−2)−7x(rxr−1) + 12xr= 0
Step 4: Simplify the equation.
r(r−1)xr−7rxr+ 12xr= 0
Step 5: Factor out xr.
xr(r2−8r+ 12) = 0
Step 6: Set each factor equal to zero and solve for r.
r2−8r+ 12 = 0
Step 7: Find the roots of the characteristic equation.
r=8±p(−8)2−4(1)(12)
2(1) =8±√16
2= 4 ±2
So, r1= 6 and r2= 2.
Step 8: Write down the general solution.
y(x) = c1x6+c2x2
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −7xy′+
12y= 0 is y(x) = c1x6+c2x2.
Question 5
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
4
Solution
Step 1: Write the auxiliary equation by assuming a solution of the form y=xr.
Step 2: Differentiate ytwice and substitute it into the differential equation.
Step 3: Simplify the resulting equation by dividing through by x2. Step 4:
Write the characteristic equation. Step 5: Solve the characteristic equation to
find the roots. Step 6: Write down the general solution for the Cauchy-Euler
equation.
Step 1: Assume y=xr.
Step 2: Differentiate ytwice:
y′=rxr−1
y′′ =r(r−1)xr−2
Substitute into the differential equation:
x2(r(r−1)xr−2)+3xr(r−1)xr−1−4xr= 0
Step 3: Divide through by x2:
r(r−1) + 3r−3−4=0
Step 4: Write the characteristic equation:
r2−r+ 3r−3−4=0
r2+ 2r−7=0
Step 5: Solve the characteristic equation to find the roots:
r=−2±p22−4(1)(−7)
2(1)
r=−2±√28
2
r=−2±2√7
2
r=−1±√7
Step 6: The general solution for the Cauchy-Euler equation is:
y(x) = c1x−1+√7+c2x−1−√7
Question 6
Question
Solve the Cauchy-Euler equation (x2−1)y′′−3xy′+4y= 0 using the substitution
y=xr.
5
Solution
Step 1: Find the derivatives of y.
y=xr⇒y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives back into the differential equation.
(x2−1)r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 3: Simplify and rewrite the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Solve the characteristic equation r2−4r+ 4 = 0.
(r−2)2= 0 ⇒r= 2 with multiplicity 2
Step 6: Write down the general solution for the differential equation.
y=c1x2+c2x2ln |x|where c1, c2∈R
Therefore, the general solution to the Cauchy-Euler equation (x2−1)y′′ −
3xy′+ 4y= 0 is y=c1x2+c2x2ln |x|.
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Rewrite the given Cauchy-Euler equation in terms of r:
r(r−1)xr−3rxr+ 4xr= 0
Step 2: Simplify the equation:
r2xr−rxr−3rxr+ 4xr= 0
r2xr−4rxr+ 4xr= 0
r2−4r+ 4 = 0
6
Step 3: Solve the quadratic equation r2−4r+ 4 = 0 to find the roots:
(r−2)2= 0
r= 2
Step 4: Write down the general solution using the roots obtained:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
Question 8
Question
Solve the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equa-
tion:
x2(r(r−1)xr−2)−x(rxr−1) + xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−rxr+xr= 0
r2xr−rxr+xr= 0
Step 5: Factor out xr:
xr(r2−r+ 1) = 0
Step 6: Set the factor equal to zero and solve for r:
r2−r+ 1 = 0
Step 7: Use the quadratic formula to find the roots of the characteristic
equation:
r=1±√1−4
2=1±i√3
2
Thus, the general solution to the Cauchy-Euler equation is
y(x) = c1x1+i√3
2+c2x1−i√3
2
7
Question 9
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ +x(x2−4)y′−6y= 0.
Solution
Step 1: First, we assume a solution in the form of y=xm, where mis a constant
to be determined.
Step 2: We differentiate y=xmtwice to find y′and y′′:
y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substituting y,y′, and y′′ into the Cauchy-Euler equation, we get:
(x2−4)x2(m(m−1)xm−2) + x(x2−4)(mxm−1)−6xm= 0.
Step 4: Simplifying the equation, we have:
(m(m−1)xm(x2−4) + mx(x2−4)xm−6xm= 0.
Step 5: Further simplification gives us:
m(m−1)(x2−4)xm+mx3(x2−4) −6xm= 0.
Step 6: Rearranging terms, we obtain:
m(m−1)xm+2 −4mxm+1 +mxm+3 −4mxm+2 −6xm= 0.
Step 7: Combine like terms and factor out xm:
m(m−1)xm+2 +mxm+3 −4mxm+2 −4mxm+1 −6xm= 0.
Step 8: Factor out xm+1 from the equation:
xm+1(mx2+ (m−4)x−6) = 0.
Step 9: Since this equation must hold for all x= 0, the polynomial inside
the parentheses must be zero:
mx2+ (m−4)x−6=0.
Step 10: Solve the quadratic equation mx2+ (m−4)x−6 = 0 for xto find
the roots.
Step 11: Once the roots of the characteristic equation are found, substitute
them back into the general solution y=Axm1+Bxm2, where m1and m2are
the roots and Aand Bare arbitrary constants.
Step 12: The final solution is the linear combination of the solutions obtained
in step 11.
8
Question 10
Question
Consider the Cauchy-Euler equation:
x2y′′ −3xy′+ 5y= 0
Find the general solution to this differential equation.
Solution
Step 1: Let’s assume a solution of the form y(x) = xrwhere ris a real number.
Step 2: Calculate the first and second derivatives of y(x):
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+5xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+ 5xr= 0
Step 5: Factor out xr:
xr(r2−4r+ 5) = 0
Step 6: Set r2−4r+ 5 = 0 and solve for rusing the quadratic formula:
r=4±p(−4)2−4(1)(5)
2(1)
r=4±√16 −20
2
r=4±√−4
2
Since the discriminant is negative, we have complex roots:
r=4±2i
2
r= 2 ±i
Step 7: The general solution to the Cauchy-Euler equation is:
y(x) = c1x2cos(x) + c2x2sin(x)
where c1and c2are arbitrary constants.
9
Question 11
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr. Then calculate the derivatives.
Let y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the original equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
Step 3: Solve the resulting characteristic equation for r.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 4: Form the general solution.
y(x) = c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln(x).
Question 12
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 3y= 0 using the substitution
y=xm.
Solution
Step 1: Let’s start by finding the first and second derivatives of y=xmwith
respect to x.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
10
Step 2: Now, substitute y,y′, and y′′ into the Cauchy-Euler equation x2y′′ −
3xy′+ 3y= 0.
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
Step 3: Simplify the equation by collecting like terms and factoring out
common factors.
m(m−1)xm−3mxm+ 3xm= 0
Step 4: Factor out xm.
xm(m2−4m+ 3) = 0
Step 5: Set each factor equal to zero to find possible values of m.
m2−4m+ 3 = 0
(m−1)(m−3) = 0
This gives m= 1 and m= 3 as the roots.
Step 6: Write down the general solution using the roots m= 1 and m= 3.
y=c1x1+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
3y= 0 is y=c1x+c2x3where c1and c2are arbitrary constants.
Question 13
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y=x2ln(x).
Solution
Step 1: Assume a solution of the form y=xm. Step 2: Calculate the first
and second derivatives of y=xm. Step 3: Substitute y=xm,y′=mxm−1,
and y′′ =m(m−1)xm−2into the Cauchy-Euler equation. Step 4: Simplify the
equation obtained in Step 3. Step 5: Equate the terms involving xmto xmln(x)
on the right side of the equation. Step 6: Solve for mby matching coefficients.
Step 7: Write the general solution using the roots found in Step 6.
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
11
Solution
Step 1: Assume the solution has the form y=xrwhere ris a constant to be
determined.
Step 2: Find the first and second derivatives of y.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(y′′)−4x(y′)+6y=x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation.
r(r−1)xr−4rxr+ 6xr= 0
Step 5: Factor out xr.
xr(r(r−1) −4r+ 6) = 0
Step 6: Set the expression in the brackets equal to zero to find possible
values for r.
r2−r−4r+ 6 = r2−5r+ 6 = 0
(r−2)(r−3) = 0
r= 2, r = 3
Step 7: Write the general solution using the roots found in Step 6.
y=c1x2+c2x3
where c1and c2are arbitrary constants.
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we first assume a
solution of the form y=xr. We then find the derivatives needed to substitute
back into the original equation.
Step 1: Find the first and second derivatives of y=xr
y′=rxr−1
12
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation
Substitute y=xr,y′, and y′′ into x2y′′ −3xy′+ 4y= 0.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Expand and simplify:
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Combine like terms Combine the terms involving xr:
r(r−1) −3r+ 4 = 0
Step 4: Solve for rSolve the quadratic equation for r:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 5: Write down the general solution Since we found r= 2, the
general solution to the Cauchy-Euler equation is given by:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 16
Question
Solve the Cauchy-Euler equation (x2−4)y′′ −xy′+y= 0 using the substitution
y=xm.
Solution
Step 1: Find the first and second derivatives of y=xm.
y=xm,
y′=mxm−1,
y′′ =m(m−1)xm−2.
13
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
(x2−4)y′′ −xy′+y= (x2−4)m(m−1)xm−2−x(mxm−1) + xm
=m(m−1)xm−4m(m−1)xm−2−mxm+xm
=m(m−1)xm−4m(m−1)xm−mxm+xm
= (m2−4m−m+ 1)xm
= (m2−5m+ 1)xm.
Step 3: Set the expression equal to zero and solve for m.
(m2−5m+ 1)xm= 0
m2−5m+ 1 = 0.
Step 4: Solve for musing the quadratic formula.
m=5±p52−4(1)(1)
2
m=5±√21
2.
Step 5: Determine the roots of the characteristic equation.
m1=5 + √21
2,
m2=5−√21
2.
Step 6: Write the general solution using the roots m1and m2.
y(x) = c1x(5+√21)/2+c2x(5−√21)/2.
Question 17
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −3xy′+y= 0.
Solution
1. Let’s first convert the given equation into standard form. Assume y=xr,
differentiating twice gives:
(x2−4)x2r(r−1)xr−2−3xr(xr−1) + xr= 0
Simplifying and combining like terms yields:
(r2−3r+ 1)xr= 0
14
2. This is the characteristic equation. Therefore, we have the roots r1and
r2where:
r1=3 + √5
2and r2=3−√5
2
3. The general solution to the Cauchy-Euler equation is given by:
y(x) = c1xr1+c2xr2
Therefore, the solution to the given Cauchy-Euler equation is:
y(x) = c1x3+√5
2+c2x3−√5
2
where c1and c2are arbitrary constants.
Question 18
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 5y= 0.
Solution
Step 1: Find the characteristic equation by substituting y=xrinto the Cauchy-
Euler equation.
x2r(r−1)xr−2−3xrxr−1+ 5xr= 0
This simplifies to:
r(r−1) −3r+ 5 = 0
r2−4r+ 5 = 0
Step 2: Solve for rby using the quadratic formula.
r=4±p(−4)2−4(1)(5)
2
r=4±√16 −20
2
r=4±√−4
2
r=4±2i
2
r= 2 ±i
Therefore, the roots are complex: r1= 2 + iand r2= 2 −i.
Step 3: Solve for the general solution using the roots found in Step 2.
y=c1x2+i+c2x2−i
15
y=c1x2(cos(1 ln x) + isin(1 ln x)) + c2x2(cos(−1 ln x) + isin(−1 ln x))
y=c1x2(cos(ln x) + isin(ln x)) + c2x2(cos(−ln x) + isin(−ln x))
y=x2(c1cos(ln x) + c2cos(ln x)) + x2i(c1sin(ln x)−c2sin(ln x))
y=x2[(c1+c2) cos(ln x) + i(c1−c2) sin(ln x)]
y=x2[Acos(ln x) + Bsin(ln x)]
where A=c1+c2and B=i(c1−c2) are constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
5y= 0 is y=x2[Acos(ln x) + Bsin(ln x)].
Question 19
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 9y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first and second derivatives of y(x).
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 3: Substitute y(x), y′(x), and y′′(x) into the differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
Step 4: Simplify the equation by expanding and combining like terms.
r(r−1)xr−3rxr+ 9xr= 0
Step 5: Factor out xrfrom the equation.
xr(r(r−1) −3r+ 9) = 0
Step 6: Set the expression inside the parentheses equal to zero.
r(r−1) −3r+ 9 = 0
Step 7: Solve the quadratic equation for r.
r2−r−3r+ 9 = 0 =⇒r2−4r+ 9 = 0 =⇒(r−3)2= 0
Step 8: Since (r−3)2= 0, we have a repeated root r= 3.
Step 9: The general solution to the Cauchy-Euler equation is given by:
y(x) = c1x3+c2x3ln(x)
where c1and c2are arbitrary constants.
16
Question 20
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
Solution
To solve the Cauchy-Euler equation, we will make the substitution y=xmand
find the values of mthat satisfy the equation.
Step 1: Find the derivatives of y=xm.
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y=xm,y′, and y′′ into the original differential equation.
x2(m(m−1)xm−2)−x(mxm−1) + xm=x3
Step 3: Simplify the equation.
m(m−1)xm−mxm+xm=x3
Step 4: Combine like terms.
(m2−m+ 1)xm=x3
Step 5: Set the equation equal to zero: m2−m+ 1 = 0. This is a quadratic
equation and the solutions can be found using the quadratic formula:
m=−(−1) ±p(−1)2−4(1)(1)
2(1)
m=1±√1−4
2
m=1±√−3
2
m=1±i√3
2
Step 6: The roots of the characteristic equation are complex. Therefore,
the general solution to the differential equation is:
y(x) = x1/2"c1cos √3 ln x
2!+c2sin √3 ln x
2!#+x3
where c1and c2are arbitrary constants.
17
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −2xy′+ 2y=x2ln(x).
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand find
y′and y′′.
Step 1: Find y′and y′′
Given y=xm, we have y′=mxm−1and y′′ =m(m−1)xm−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation
The Cauchy-Euler equation becomes: x2(m)(m−1)xm−2−2x(mxm−1) +
2xm=x2ln(x).
Simplifying, we get: m(m−1)xm−2mxm+ 2xm=x2ln(x).
This simplifies to the following differential equation: m2−3m+2 = ln(x).
Step 3: Solve the characteristic equation
To solve for m, we set the characteristic equation m2−3m+ 2 = 0.
Factoring, we get (m−1)(m−2) = 0.
So, m= 1 or m= 2.
Step 4: Find the general solution
Since we have repeated roots, the general solution is: y(x) = c1x+c2x2.
Step 5: Find the particular solution for ln(x)
Since the right-hand side is ln(x), we try a particular solution of the form
yp=Aln(x).
Determine Aby substituting ypinto the differential equation.
Find y′
pand y′′
p, then substitute into the equation.
After solving for A, we get A=−1
2.
Step 6: Combine the general and particular solutions
The general solution is y(x) = c1x+c2x2−1
2xln(x).
Therefore, the solution to the Cauchy-Euler equation x2y′′ −2xy′+ 2y=
x2ln(x) is y(x) = c1x+c2x2−1
2xln(x).
18
Question 22
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate to find y′
and y′′. Step 2: Substitute y,y′, and y′′ back into the differential equation.
Step 3: Solve the resulting equation obtained in step 2 to find the characteristic
equation. Step 4: Find the roots of the characteristic equation. Step 5: Form
the general solution using the roots obtained in step 4. Step 6: State the final
solution by writing out the complete expression for y(x).
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
Step 1: Find the derivatives of y=xm.
y=xm⇒y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute y=xm,y′, and y′′ into the Cauchy-Euler equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify the equation.
m(m−1)xm−3mxm+ 4xm= 0
Step 4: Factor out xm.
xm(m(m−1) −3m+ 4) = 0
Step 5: Solve the characteristic equation.
m2−m−3m+ 4 = 0 ⇒m2−4m+ 4 = 0 ⇒(m−2)2= 0
Step 6: Determine the roots of the characteristic equation.
m= 2 (repeated root)
Step 7: Write down the general solution.
y=c1x2+c2x2ln(x)
Therefore, the solution to the Cauchy-Euler equation is y=c1x2+c2x2ln(x).
19
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y=xr.
Step 2: Find the first and second derivatives of y:
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−4rxr+ 6xr= 0
Step 5: Factor out xr:
xr(r2−5r+ 6) = 0
Step 6: Find the roots of the characteristic equation r2−5r+ 6 = 0:
r2−5r+ 6 = (r−2)(r−3) = 0
r1= 2 and r2= 3
Step 7: Write the general solution using the roots r1and r2:
y(x) = c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y(x) = c1x2+c2x3, where c1and c2are constants.
Question 25
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −4xy′+ 6y= 0 using the substi-
tution y=xr.
20
Solution
Step 1: Find the derivatives of y=xrwith respect to x.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equation.
(x2−4)x2y′′ −4xy′+ 6y= 0
(x2−4)x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
(x2−4)r(r−1)xr−4rxr+ 6xr= 0
Step 3: Simplify the equation.
(x2−4)r(r−1)xr−4rxr+ 6xr= 0
r(r−1)xr+2 −4rxr+1 + 6xr+2 = 0
r(r−1)xr+2 −4rxr+1 + 6xr+2 = 0
Step 4: Divide through by xr+2 to get a quadratic equation in r.
r(r−1) −4r+ 6 = 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
(r−2)(r−3) = 0
Step 5: Find the roots of the quadratic equation.
r1= 2, r2= 3
Step 6: Write down the two linearly independent solutions from the roots
found. y1=x2
y2=x3
Therefore, the general solution to the Cauchy-Euler equation is y=c1x2+
c2x3, where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −y= 0 using the substitution y=xm.
Hence, find the general solution to the equation.
21
Solution
Step 1: Find the first and second derivatives of y=xm.
dy
dx =mxm−1,
d2y
dx2=m(m−1)xm−2.
Step 2: Substitute y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
given Cauchy-Euler equation x2y′′ −y= 0.
x2·m(m−1)xm−2−xm= 0.
Step 3: Simplify the equation and solve for m.
m(m−1)xm−xm= 0 =⇒m(m−1) −1 = 0 =⇒m2−m−1=0.
Step 4: Solve for musing the quadratic formula: m=−(−1)±√(−1)2−4(1)(−1)
2(1) .
m=1±√1+4
2=1±√5
2.
Step 5: Now, we have two roots m1=1+√5
2and m2=1−√5
2. This gives
two linearly independent solutions xm1and xm2.
Step 6: Therefore, the general solution to the Cauchy-Euler equation x2y′′ −
y= 0 is y=C1x(1+√5)/2+C2x(1−√5)/2, where C1and C2are arbitrary con-
stants.
Question 27
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Let’s first find the characteristic equation by substituting y=xrinto
the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
We have:
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Substitute these expressions into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
22
r(r−1)xr−3rxr+ 4xr=x3
r2−4r+ 4 = 1
r2−4r+ 3 = 0
(r−1)(r−3) = 0
So, we have two roots r= 1 and r= 3.
Step 2: Now that we have the roots of the characteristic equation, we can
write the general solution. The general solution will be:
y=c1x1+c2x3
y=c1x+c2x3
Step 3: Let’s now find the particular solution for the nonhomogeneous term
x3. We try a particular solution of the form yp=Ax3.
y′
p= 3Ax2
y′′
p= 6Ax
Substitute these into the Cauchy-Euler equation:
x2(6Ax)−3x(3Ax2) + 4(Ax3) = x3
6Ax3−9Ax3+ 4Ax3=x3
Ax3=x3
A= 1
So, the particular solution is yp=x3.
Step 4: The general solution is the sum of the homogeneous and particular
solutions.
y=c1x+c2x3+x3
y=c1x+c2x3+x3
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 3y= 0
23
Solution
Step 1: Write the characteristic equation by assuming y=xm. Step 2: Find
the roots of the characteristic equation. Step 3: Form the general solution.
Step 1: Assume y=xm, then differentiate to find y′and y′′.
y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the differential equation and
simplify.
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
m(m−1)xm−3mxm+ 3xm= 0
m2−4m+ 3 = m(m−3)(m−1) = 0
The roots are m= 1,3. So, the general solution is
y(x) = c1x+c2x3
Question 29
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
Solution
Step 1: Find the characteristic equation by substituting y=xminto the Cauchy-
Euler equation. The characteristic equation is obtained by solving the following
equation:
m(m−1)xm−2−mxm−1+xm= 0
Step 2: Simplify the characteristic equation to find the roots m1and m2.
m(m−1) −m+ 1 = m2−m−m+ 1 = m2−2m+ 1 = (m−1)2= 0
This gives m1=m2= 1.
Step 3: Write down the general solution using the roots m1and m2.
y(x) = c1xm1+c2xm2=c1x+c2x
Step 4: Find a particular solution to the Cauchy-Euler equation for the part
x3by assuming yp=Ax3and substituting into the equation.
2A−3Ax +Ax =x3
2A= 1
A=1
2
24
Step 5: Write down the general solution as the sum of the general solution
and the particular solution.
y(x) = c1x+c2x+1
2x3
Therefore, the solution to the Cauchy-Euler equation x2y′′ −xy′+y=x3is
y(x) = c1x+c2x+1
2x3.
Question 30
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Begin by assuming a solution of the form y=xr.
Step 2: Calculate the first and second derivatives of y:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+ 4xr= 0
Step 5: Factor out xr:
xr(r(r−1) −3r+ 4) = 0
Step 6: Find the roots of the characteristic equation r(r−1) −3r+ 4 = 0.
Step 7: Solve the quadratic equation r2−r−3r+ 4 = 0.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 8: Since r= 2 is a repeated root, the general solution is y=c1x2+
c2x2ln x, where c1and c2are arbitrary constants.
25
Question 31
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.Then find the derivatives of
y.
y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y, y′,and y′′ into the differential equation.
x2y′′ + 3xy′−4y=x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 3: Simplify the equation obtained in Step 2.
r(r−1)xr+ 3rxr−4xr=xr[r(r−1) + 3r−4] = 0
Step 4: Set the expression in square brackets equal to zero and solve for r.
r(r−1) + 3r−4 = r2−r+ 3r−4 = r2+ 2r−4=0
r2+ 2r−4=(r+ 2)(r−2) = 0
Thus, r=−2 or r= 2.
Step 5: Determine the general solution using the roots obtained in Step 4.
The general solution is:
y(x) = c1x−2+c2x2
where c1and c2are arbitrary constants.
Question 32
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
26
Solution
Step 1: Write down the characteristic equation of the Cauchy-Euler equation
by substituting y=xr:
r(r−1)xr−5rxr+ 6xr= 0
Step 2: Simplify the equation by factoring out xr:
xr(r2−6r+ 6) = 0
Step 3: Solve the characteristic equation r2−6r+ 6 = 0 for the roots r:
r=6±p(−6)2−4(1)(6)
2(1) =6±√36 −24
2=6±√12
2= 3 ±√3
Step 4: Write down the general solution to the Cauchy-Euler equation using
the roots:
y(x) = c1x3+√3+c2x3−√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y= 0 is y(x) = c1x3+√3+c2x3−√3.
Question 33
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ + 4xy′−6y= 0.
Solution
Step 1: Write the equation in standard Cauchy-Euler form by assuming y=xr.
(x2−4)x2y′′ + 4xy′−6y= 0
⇒(x2−4)x2(r)(r−1)xr−2+ 4x(r)xr−1−6xr= 0
⇒(r)(r−1)xr−4(r)xr−6xr= 0
⇒r2−5r−6=0
Step 2: Solve the quadratic equation r2−5r−6 = 0 to find the roots r1and
r2.
r1=5 + p52−4(1)(−6)
2(1)
=5 + √25 + 24
2
=5 + √49
2
=5+7
2
= 6
27
r2=5−p52−4(1)(−6)
2(1)
=5−√25 + 24
2
=5−√49
2
=5−7
2
=−1
Step 3: Write the general solution using the roots r1and r2.
y(x) = c1xr1+c2xr2
=c1x6+c2x−1
where c1and c2are arbitrary constants.
Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 12y= 0.
Solution
To solve the Cauchy-Euler equation, we first make the substitution y=xm
where mis a constant to be determined.
Step 1: Find the derivatives of y.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives back into the equation. The Cauchy-
Euler equation becomes:
x2(m(m−1)xm−2)−5x(mxm−1) + 12xm= 0
Step 3: Simplify the equation.
m(m−1)xm−5mxm+ 12xm= 0
m2−m−5m+ 12 = 0
m2−6m+ 12 = 0
28
Step 4: Solve the characteristic equation. We solve the quadratic equation
m2−6m+ 12 = 0 using the quadratic formula:
m=−(−6) ±p(−6)2−4(1)(12)
2(1)
m=6±√36 −48
2
m=6±√−12
2
m=6±2i√3
2
m= 3 ±i√3
Step 5: Determine the general solution. Therefore, the general solution to
the Cauchy-Euler equation is:
y=c1x3+i√3+c2x3−i√3
or
y=c1x3(cos√3 ln(x)+isin√3 ln(x))+c2x3(cos√3 ln(x)−isin√3 ln(x))
Question 35
Question
Solve the following Cauchy-Euler differential equation:
x2y′′ −3xy′+ 3y=x3
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the given differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+3xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 3xr=x3
29
Question 2
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
To solve the given Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we will assume
a solution of the form y=xrand then find the values of r.
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equa-
tion.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
Step 4: Solve the resulting quadratic equation for r.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 5: Form the general solution. Since r= 2, the solution to the
Cauchy-Euler equation is y=c1x2+c2x2ln x, where c1and c2are arbitrary
constants.
Question 3
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
2
Solution
Step 1: Assume a solution of the form y=xrwhere ris a constant to be
determined. Then, calculate the first and second derivatives of y.
Let y=xr
First derivative: y′=rxr−1
Second derivative: y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation, and simplify.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out an xrfrom the equation.
xr(r(r−1) −3r+ 4) = 0
Step 4: Set the expression inside the parentheses equal to zero to find the
characteristic equation.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 5: Find the roots of the characteristic equation.
r= 2
Step 6: Form the general solution using the roots found in Step 5.
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x), where c1and c2are arbitrary constants.
Question 4
Question
Solve the Cauchy-Euler equation x2y′′ −7xy′+ 12y= 0.
3
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Calculate the first and second derivatives of y.
y′=rxr−1, y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the original differential equation.
x2(r(r−1)xr−2)−7x(rxr−1) + 12xr= 0
Step 4: Simplify the equation.
r(r−1)xr−7rxr+ 12xr= 0
Step 5: Factor out xr.
xr(r2−8r+ 12) = 0
Step 6: Set each factor equal to zero and solve for r.
r2−8r+ 12 = 0
Step 7: Find the roots of the characteristic equation.
r=8±p(−8)2−4(1)(12)
2(1) =8±√16
2= 4 ±2
So, r1= 6 and r2= 2.
Step 8: Write down the general solution.
y(x) = c1x6+c2x2
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −7xy′+
12y= 0 is y(x) = c1x6+c2x2.
Question 5
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
4
Solution
Step 1: Write the auxiliary equation by assuming a solution of the form y=xr.
Step 2: Differentiate ytwice and substitute it into the differential equation.
Step 3: Simplify the resulting equation by dividing through by x2. Step 4:
Write the characteristic equation. Step 5: Solve the characteristic equation to
find the roots. Step 6: Write down the general solution for the Cauchy-Euler
equation.
Step 1: Assume y=xr.
Step 2: Differentiate ytwice:
y′=rxr−1
y′′ =r(r−1)xr−2
Substitute into the differential equation:
x2(r(r−1)xr−2)+3xr(r−1)xr−1−4xr= 0
Step 3: Divide through by x2:
r(r−1) + 3r−3−4=0
Step 4: Write the characteristic equation:
r2−r+ 3r−3−4=0
r2+ 2r−7=0
Step 5: Solve the characteristic equation to find the roots:
r=−2±p22−4(1)(−7)
2(1)
r=−2±√28
2
r=−2±2√7
2
r=−1±√7
Step 6: The general solution for the Cauchy-Euler equation is:
y(x) = c1x−1+√7+c2x−1−√7
Question 6
Question
Solve the Cauchy-Euler equation (x2−1)y′′−3xy′+4y= 0 using the substitution
y=xr.
5
Solution
Step 1: Find the derivatives of y.
y=xr⇒y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives back into the differential equation.
(x2−1)r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 3: Simplify and rewrite the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Solve the characteristic equation r2−4r+ 4 = 0.
(r−2)2= 0 ⇒r= 2 with multiplicity 2
Step 6: Write down the general solution for the differential equation.
y=c1x2+c2x2ln |x|where c1, c2∈R
Therefore, the general solution to the Cauchy-Euler equation (x2−1)y′′ −
3xy′+ 4y= 0 is y=c1x2+c2x2ln |x|.
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Rewrite the given Cauchy-Euler equation in terms of r:
r(r−1)xr−3rxr+ 4xr= 0
Step 2: Simplify the equation:
r2xr−rxr−3rxr+ 4xr= 0
r2xr−4rxr+ 4xr= 0
r2−4r+ 4 = 0
6
Step 3: Solve the quadratic equation r2−4r+ 4 = 0 to find the roots:
(r−2)2= 0
r= 2
Step 4: Write down the general solution using the roots obtained:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
Question 8
Question
Solve the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equa-
tion:
x2(r(r−1)xr−2)−x(rxr−1) + xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−rxr+xr= 0
r2xr−rxr+xr= 0
Step 5: Factor out xr:
xr(r2−r+ 1) = 0
Step 6: Set the factor equal to zero and solve for r:
r2−r+ 1 = 0
Step 7: Use the quadratic formula to find the roots of the characteristic
equation:
r=1±√1−4
2=1±i√3
2
Thus, the general solution to the Cauchy-Euler equation is
y(x) = c1x1+i√3
2+c2x1−i√3
2
7
Question 9
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ +x(x2−4)y′−6y= 0.
Solution
Step 1: First, we assume a solution in the form of y=xm, where mis a constant
to be determined.
Step 2: We differentiate y=xmtwice to find y′and y′′:
y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substituting y,y′, and y′′ into the Cauchy-Euler equation, we get:
(x2−4)x2(m(m−1)xm−2) + x(x2−4)(mxm−1)−6xm= 0.
Step 4: Simplifying the equation, we have:
(m(m−1)xm(x2−4) + mx(x2−4)xm−6xm= 0.
Step 5: Further simplification gives us:
m(m−1)(x2−4)xm+mx3(x2−4) −6xm= 0.
Step 6: Rearranging terms, we obtain:
m(m−1)xm+2 −4mxm+1 +mxm+3 −4mxm+2 −6xm= 0.
Step 7: Combine like terms and factor out xm:
m(m−1)xm+2 +mxm+3 −4mxm+2 −4mxm+1 −6xm= 0.
Step 8: Factor out xm+1 from the equation:
xm+1(mx2+ (m−4)x−6) = 0.
Step 9: Since this equation must hold for all x= 0, the polynomial inside
the parentheses must be zero:
mx2+ (m−4)x−6=0.
Step 10: Solve the quadratic equation mx2+ (m−4)x−6 = 0 for xto find
the roots.
Step 11: Once the roots of the characteristic equation are found, substitute
them back into the general solution y=Axm1+Bxm2, where m1and m2are
the roots and Aand Bare arbitrary constants.
Step 12: The final solution is the linear combination of the solutions obtained
in step 11.
8
Question 10
Question
Consider the Cauchy-Euler equation:
x2y′′ −3xy′+ 5y= 0
Find the general solution to this differential equation.
Solution
Step 1: Let’s assume a solution of the form y(x) = xrwhere ris a real number.
Step 2: Calculate the first and second derivatives of y(x):
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+5xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+ 5xr= 0
Step 5: Factor out xr:
xr(r2−4r+ 5) = 0
Step 6: Set r2−4r+ 5 = 0 and solve for rusing the quadratic formula:
r=4±p(−4)2−4(1)(5)
2(1)
r=4±√16 −20
2
r=4±√−4
2
Since the discriminant is negative, we have complex roots:
r=4±2i
2
r= 2 ±i
Step 7: The general solution to the Cauchy-Euler equation is:
y(x) = c1x2cos(x) + c2x2sin(x)
where c1and c2are arbitrary constants.
9
Question 11
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr. Then calculate the derivatives.
Let y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the original equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
Step 3: Solve the resulting characteristic equation for r.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 4: Form the general solution.
y(x) = c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln(x).
Question 12
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 3y= 0 using the substitution
y=xm.
Solution
Step 1: Let’s start by finding the first and second derivatives of y=xmwith
respect to x.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
10
Step 2: Now, substitute y,y′, and y′′ into the Cauchy-Euler equation x2y′′ −
3xy′+ 3y= 0.
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
Step 3: Simplify the equation by collecting like terms and factoring out
common factors.
m(m−1)xm−3mxm+ 3xm= 0
Step 4: Factor out xm.
xm(m2−4m+ 3) = 0
Step 5: Set each factor equal to zero to find possible values of m.
m2−4m+ 3 = 0
(m−1)(m−3) = 0
This gives m= 1 and m= 3 as the roots.
Step 6: Write down the general solution using the roots m= 1 and m= 3.
y=c1x1+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
3y= 0 is y=c1x+c2x3where c1and c2are arbitrary constants.
Question 13
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y=x2ln(x).
Solution
Step 1: Assume a solution of the form y=xm. Step 2: Calculate the first
and second derivatives of y=xm. Step 3: Substitute y=xm,y′=mxm−1,
and y′′ =m(m−1)xm−2into the Cauchy-Euler equation. Step 4: Simplify the
equation obtained in Step 3. Step 5: Equate the terms involving xmto xmln(x)
on the right side of the equation. Step 6: Solve for mby matching coefficients.
Step 7: Write the general solution using the roots found in Step 6.
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
11
Solution
Step 1: Assume the solution has the form y=xrwhere ris a constant to be
determined.
Step 2: Find the first and second derivatives of y.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(y′′)−4x(y′)+6y=x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation.
r(r−1)xr−4rxr+ 6xr= 0
Step 5: Factor out xr.
xr(r(r−1) −4r+ 6) = 0
Step 6: Set the expression in the brackets equal to zero to find possible
values for r.
r2−r−4r+ 6 = r2−5r+ 6 = 0
(r−2)(r−3) = 0
r= 2, r = 3
Step 7: Write the general solution using the roots found in Step 6.
y=c1x2+c2x3
where c1and c2are arbitrary constants.
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we first assume a
solution of the form y=xr. We then find the derivatives needed to substitute
back into the original equation.
Step 1: Find the first and second derivatives of y=xr
y′=rxr−1
12
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation
Substitute y=xr,y′, and y′′ into x2y′′ −3xy′+ 4y= 0.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Expand and simplify:
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Combine like terms Combine the terms involving xr:
r(r−1) −3r+ 4 = 0
Step 4: Solve for rSolve the quadratic equation for r:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 5: Write down the general solution Since we found r= 2, the
general solution to the Cauchy-Euler equation is given by:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 16
Question
Solve the Cauchy-Euler equation (x2−4)y′′ −xy′+y= 0 using the substitution
y=xm.
Solution
Step 1: Find the first and second derivatives of y=xm.
y=xm,
y′=mxm−1,
y′′ =m(m−1)xm−2.
13
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
(x2−4)y′′ −xy′+y= (x2−4)m(m−1)xm−2−x(mxm−1) + xm
=m(m−1)xm−4m(m−1)xm−2−mxm+xm
=m(m−1)xm−4m(m−1)xm−mxm+xm
= (m2−4m−m+ 1)xm
= (m2−5m+ 1)xm.
Step 3: Set the expression equal to zero and solve for m.
(m2−5m+ 1)xm= 0
m2−5m+ 1 = 0.
Step 4: Solve for musing the quadratic formula.
m=5±p52−4(1)(1)
2
m=5±√21
2.
Step 5: Determine the roots of the characteristic equation.
m1=5 + √21
2,
m2=5−√21
2.
Step 6: Write the general solution using the roots m1and m2.
y(x) = c1x(5+√21)/2+c2x(5−√21)/2.
Question 17
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −3xy′+y= 0.
Solution
1. Let’s first convert the given equation into standard form. Assume y=xr,
differentiating twice gives:
(x2−4)x2r(r−1)xr−2−3xr(xr−1) + xr= 0
Simplifying and combining like terms yields:
(r2−3r+ 1)xr= 0
14
2. This is the characteristic equation. Therefore, we have the roots r1and
r2where:
r1=3 + √5
2and r2=3−√5
2
3. The general solution to the Cauchy-Euler equation is given by:
y(x) = c1xr1+c2xr2
Therefore, the solution to the given Cauchy-Euler equation is:
y(x) = c1x3+√5
2+c2x3−√5
2
where c1and c2are arbitrary constants.
Question 18
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 5y= 0.
Solution
Step 1: Find the characteristic equation by substituting y=xrinto the Cauchy-
Euler equation.
x2r(r−1)xr−2−3xrxr−1+ 5xr= 0
This simplifies to:
r(r−1) −3r+ 5 = 0
r2−4r+ 5 = 0
Step 2: Solve for rby using the quadratic formula.
r=4±p(−4)2−4(1)(5)
2
r=4±√16 −20
2
r=4±√−4
2
r=4±2i
2
r= 2 ±i
Therefore, the roots are complex: r1= 2 + iand r2= 2 −i.
Step 3: Solve for the general solution using the roots found in Step 2.
y=c1x2+i+c2x2−i
15
y=c1x2(cos(1 ln x) + isin(1 ln x)) + c2x2(cos(−1 ln x) + isin(−1 ln x))
y=c1x2(cos(ln x) + isin(ln x)) + c2x2(cos(−ln x) + isin(−ln x))
y=x2(c1cos(ln x) + c2cos(ln x)) + x2i(c1sin(ln x)−c2sin(ln x))
y=x2[(c1+c2) cos(ln x) + i(c1−c2) sin(ln x)]
y=x2[Acos(ln x) + Bsin(ln x)]
where A=c1+c2and B=i(c1−c2) are constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
5y= 0 is y=x2[Acos(ln x) + Bsin(ln x)].
Question 19
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 9y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first and second derivatives of y(x).
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 3: Substitute y(x), y′(x), and y′′(x) into the differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
Step 4: Simplify the equation by expanding and combining like terms.
r(r−1)xr−3rxr+ 9xr= 0
Step 5: Factor out xrfrom the equation.
xr(r(r−1) −3r+ 9) = 0
Step 6: Set the expression inside the parentheses equal to zero.
r(r−1) −3r+ 9 = 0
Step 7: Solve the quadratic equation for r.
r2−r−3r+ 9 = 0 =⇒r2−4r+ 9 = 0 =⇒(r−3)2= 0
Step 8: Since (r−3)2= 0, we have a repeated root r= 3.
Step 9: The general solution to the Cauchy-Euler equation is given by:
y(x) = c1x3+c2x3ln(x)
where c1and c2are arbitrary constants.
16
Question 20
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
Solution
To solve the Cauchy-Euler equation, we will make the substitution y=xmand
find the values of mthat satisfy the equation.
Step 1: Find the derivatives of y=xm.
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y=xm,y′, and y′′ into the original differential equation.
x2(m(m−1)xm−2)−x(mxm−1) + xm=x3
Step 3: Simplify the equation.
m(m−1)xm−mxm+xm=x3
Step 4: Combine like terms.
(m2−m+ 1)xm=x3
Step 5: Set the equation equal to zero: m2−m+ 1 = 0. This is a quadratic
equation and the solutions can be found using the quadratic formula:
m=−(−1) ±p(−1)2−4(1)(1)
2(1)
m=1±√1−4
2
m=1±√−3
2
m=1±i√3
2
Step 6: The roots of the characteristic equation are complex. Therefore,
the general solution to the differential equation is:
y(x) = x1/2"c1cos √3 ln x
2!+c2sin √3 ln x
2!#+x3
where c1and c2are arbitrary constants.
17
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −2xy′+ 2y=x2ln(x).
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand find
y′and y′′.
Step 1: Find y′and y′′
Given y=xm, we have y′=mxm−1and y′′ =m(m−1)xm−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation
The Cauchy-Euler equation becomes: x2(m)(m−1)xm−2−2x(mxm−1) +
2xm=x2ln(x).
Simplifying, we get: m(m−1)xm−2mxm+ 2xm=x2ln(x).
This simplifies to the following differential equation: m2−3m+ 2 = ln(x).
Step 3: Solve the characteristic equation
To solve for m, we set the characteristic equation m2−3m+ 2 = 0.
Factoring, we get (m−1)(m−2) = 0.
So, m= 1 or m= 2.
Step 4: Find the general solution
Since we have repeated roots, the general solution is: y(x) = c1x+c2x2.
Step 5: Find the particular solution for ln(x)
Since the right-hand side is ln(x), we try a particular solution of the form
yp=Aln(x).
Determine Aby substituting ypinto the differential equation.
Find y′
pand y′′
p, then substitute into the equation.
After solving for A, we get A=−1
2.
Step 6: Combine the general and particular solutions
The general solution is y(x) = c1x+c2x2−1
2xln(x).
Therefore, the solution to the Cauchy-Euler equation x2y′′ −2xy′+ 2y=
x2ln(x) is y(x) = c1x+c2x2−1
2xln(x).
18
Question 22
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate to find y′
and y′′. Step 2: Substitute y,y′, and y′′ back into the differential equation.
Step 3: Solve the resulting equation obtained in step 2 to find the characteristic
equation. Step 4: Find the roots of the characteristic equation. Step 5: Form
the general solution using the roots obtained in step 4. Step 6: State the final
solution by writing out the complete expression for y(x).
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
Step 1: Find the derivatives of y=xm.
y=xm⇒y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute y=xm,y′, and y′′ into the Cauchy-Euler equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify the equation.
m(m−1)xm−3mxm+ 4xm= 0
Step 4: Factor out xm.
xm(m(m−1) −3m+ 4) = 0
Step 5: Solve the characteristic equation.
m2−m−3m+ 4 = 0 ⇒m2−4m+ 4 = 0 ⇒(m−2)2= 0
Step 6: Determine the roots of the characteristic equation.
m= 2 (repeated root)
Step 7: Write down the general solution.
y=c1x2+c2x2ln(x)
Therefore, the solution to the Cauchy-Euler equation is y=c1x2+c2x2ln(x).
19
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y=xr.
Step 2: Find the first and second derivatives of y:
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−4rxr+ 6xr= 0
Step 5: Factor out xr:
xr(r2−5r+ 6) = 0
Step 6: Find the roots of the characteristic equation r2−5r+ 6 = 0:
r2−5r+ 6 = (r−2)(r−3) = 0
r1= 2 and r2= 3
Step 7: Write the general solution using the roots r1and r2:
y(x) = c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y(x) = c1x2+c2x3, where c1and c2are constants.
Question 25
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −4xy′+ 6y= 0 using the substi-
tution y=xr.
20
Solution
Step 1: Find the derivatives of y=xrwith respect to x.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equation.
(x2−4)x2y′′ −4xy′+ 6y= 0
(x2−4)x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
(x2−4)r(r−1)xr−4rxr+ 6xr= 0
Step 3: Simplify the equation.
(x2−4)r(r−1)xr−4rxr+ 6xr= 0
r(r−1)xr+2 −4rxr+1 + 6xr+2 = 0
r(r−1)xr+2 −4rxr+1 + 6xr+2 = 0
Step 4: Divide through by xr+2 to get a quadratic equation in r.
r(r−1) −4r+ 6 = 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
(r−2)(r−3) = 0
Step 5: Find the roots of the quadratic equation.
r1= 2, r2= 3
Step 6: Write down the two linearly independent solutions from the roots
found. y1=x2
y2=x3
Therefore, the general solution to the Cauchy-Euler equation is y=c1x2+
c2x3, where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −y= 0 using the substitution y=xm.
Hence, find the general solution to the equation.
21
Solution
Step 1: Find the first and second derivatives of y=xm.
dy
dx =mxm−1,
d2y
dx2=m(m−1)xm−2.
Step 2: Substitute y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
given Cauchy-Euler equation x2y′′ −y= 0.
x2·m(m−1)xm−2−xm= 0.
Step 3: Simplify the equation and solve for m.
m(m−1)xm−xm= 0 =⇒m(m−1) −1 = 0 =⇒m2−m−1=0.
Step 4: Solve for musing the quadratic formula: m=−(−1)±√(−1)2−4(1)(−1)
2(1) .
m=1±√1+4
2=1±√5
2.
Step 5: Now, we have two roots m1=1+√5
2and m2=1−√5
2. This gives
two linearly independent solutions xm1and xm2.
Step 6: Therefore, the general solution to the Cauchy-Euler equation x2y′′ −
y= 0 is y=C1x(1+√5)/2+C2x(1−√5)/2, where C1and C2are arbitrary con-
stants.
Question 27
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Let’s first find the characteristic equation by substituting y=xrinto
the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
We have:
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Substitute these expressions into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
22
r(r−1)xr−3rxr+ 4xr=x3
r2−4r+ 4 = 1
r2−4r+ 3 = 0
(r−1)(r−3) = 0
So, we have two roots r= 1 and r= 3.
Step 2: Now that we have the roots of the characteristic equation, we can
write the general solution. The general solution will be:
y=c1x1+c2x3
y=c1x+c2x3
Step 3: Let’s now find the particular solution for the nonhomogeneous term
x3. We try a particular solution of the form yp=Ax3.
y′
p= 3Ax2
y′′
p= 6Ax
Substitute these into the Cauchy-Euler equation:
x2(6Ax)−3x(3Ax2) + 4(Ax3) = x3
6Ax3−9Ax3+ 4Ax3=x3
Ax3=x3
A= 1
So, the particular solution is yp=x3.
Step 4: The general solution is the sum of the homogeneous and particular
solutions.
y=c1x+c2x3+x3
y=c1x+c2x3+x3
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 3y= 0
23
Solution
Step 1: Write the characteristic equation by assuming y=xm. Step 2: Find
the roots of the characteristic equation. Step 3: Form the general solution.
Step 1: Assume y=xm, then differentiate to find y′and y′′.
y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the differential equation and
simplify.
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
m(m−1)xm−3mxm+ 3xm= 0
m2−4m+ 3 = m(m−3)(m−1) = 0
The roots are m= 1,3. So, the general solution is
y(x) = c1x+c2x3
Question 29
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
Solution
Step 1: Find the characteristic equation by substituting y=xminto the Cauchy-
Euler equation. The characteristic equation is obtained by solving the following
equation:
m(m−1)xm−2−mxm−1+xm= 0
Step 2: Simplify the characteristic equation to find the roots m1and m2.
m(m−1) −m+ 1 = m2−m−m+ 1 = m2−2m+ 1 = (m−1)2= 0
This gives m1=m2= 1.
Step 3: Write down the general solution using the roots m1and m2.
y(x) = c1xm1+c2xm2=c1x+c2x
Step 4: Find a particular solution to the Cauchy-Euler equation for the part
x3by assuming yp=Ax3and substituting into the equation.
2A−3Ax +Ax =x3
2A= 1
A=1
2
24
Step 5: Write down the general solution as the sum of the general solution
and the particular solution.
y(x) = c1x+c2x+1
2x3
Therefore, the solution to the Cauchy-Euler equation x2y′′ −xy′+y=x3is
y(x) = c1x+c2x+1
2x3.
Question 30
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Begin by assuming a solution of the form y=xr.
Step 2: Calculate the first and second derivatives of y:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+ 4xr= 0
Step 5: Factor out xr:
xr(r(r−1) −3r+ 4) = 0
Step 6: Find the roots of the characteristic equation r(r−1) −3r+ 4 = 0.
Step 7: Solve the quadratic equation r2−r−3r+ 4 = 0.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 8: Since r= 2 is a repeated root, the general solution is y=c1x2+
c2x2ln x, where c1and c2are arbitrary constants.
25
Question 31
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.Then find the derivatives of
y.
y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y, y′,and y′′ into the differential equation.
x2y′′ + 3xy′−4y=x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 3: Simplify the equation obtained in Step 2.
r(r−1)xr+ 3rxr−4xr=xr[r(r−1) + 3r−4] = 0
Step 4: Set the expression in square brackets equal to zero and solve for r.
r(r−1) + 3r−4 = r2−r+ 3r−4 = r2+ 2r−4=0
r2+ 2r−4=(r+ 2)(r−2) = 0
Thus, r=−2 or r= 2.
Step 5: Determine the general solution using the roots obtained in Step 4.
The general solution is:
y(x) = c1x−2+c2x2
where c1and c2are arbitrary constants.
Question 32
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
26
Solution
Step 1: Write down the characteristic equation of the Cauchy-Euler equation
by substituting y=xr:
r(r−1)xr−5rxr+ 6xr= 0
Step 2: Simplify the equation by factoring out xr:
xr(r2−6r+ 6) = 0
Step 3: Solve the characteristic equation r2−6r+ 6 = 0 for the roots r:
r=6±p(−6)2−4(1)(6)
2(1) =6±√36 −24
2=6±√12
2= 3 ±√3
Step 4: Write down the general solution to the Cauchy-Euler equation using
the roots:
y(x) = c1x3+√3+c2x3−√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y= 0 is y(x) = c1x3+√3+c2x3−√3.
Question 33
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ + 4xy′−6y= 0.
Solution
Step 1: Write the equation in standard Cauchy-Euler form by assuming y=xr.
(x2−4)x2y′′ + 4xy′−6y= 0
⇒(x2−4)x2(r)(r−1)xr−2+ 4x(r)xr−1−6xr= 0
⇒(r)(r−1)xr−4(r)xr−6xr= 0
⇒r2−5r−6=0
Step 2: Solve the quadratic equation r2−5r−6 = 0 to find the roots r1and
r2.
r1=5 + p52−4(1)(−6)
2(1)
=5 + √25 + 24
2
=5 + √49
2
=5+7
2
= 6
27
r2=5−p52−4(1)(−6)
2(1)
=5−√25 + 24
2
=5−√49
2
=5−7
2
=−1
Step 3: Write the general solution using the roots r1and r2.
y(x) = c1xr1+c2xr2
=c1x6+c2x−1
where c1and c2are arbitrary constants.
Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 12y= 0.
Solution
To solve the Cauchy-Euler equation, we first make the substitution y=xm
where mis a constant to be determined.
Step 1: Find the derivatives of y.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives back into the equation. The Cauchy-
Euler equation becomes:
x2(m(m−1)xm−2)−5x(mxm−1) + 12xm= 0
Step 3: Simplify the equation.
m(m−1)xm−5mxm+ 12xm= 0
m2−m−5m+ 12 = 0
m2−6m+ 12 = 0
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Step 4: Solve the characteristic equation. We solve the quadratic equation
m2−6m+ 12 = 0 using the quadratic formula:
m=−(−6) ±p(−6)2−4(1)(12)
2(1)
m=6±√36 −48
2
m=6±√−12
2
m=6±2i√3
2
m= 3 ±i√3
Step 5: Determine the general solution. Therefore, the general solution to
the Cauchy-Euler equation is:
y=c1x3+i√3+c2x3−i√3
or
y=c1x3(cos√3 ln(x)+isin√3 ln(x))+c2x3(cos√3 ln(x)−isin√3 ln(x))
Question 35
Question
Solve the following Cauchy-Euler differential equation:
x2y′′ −3xy′+ 3y=x3
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the given differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+3xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 3xr=x3
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Step 5: Factor out xron the left side of the equation.
xr(r(r−1) −3r+ 3) = x3
Step 6: Set the expression in the parentheses equal to 0 to find the charac-
teristic equation.
r2−4r+ 3 = 0
Step 7: Solve the characteristic equation to find the roots.
r=4±√42−4·3
2=4±√4
2= 3 or 1
Step 8: Write the general solution by combining the solutions corresponding
to the roots.
y(x) = c1x+c2x3where c1, c2are arbitrary constants
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