Assignment Section 1.3 Classification of Differential Equations
Problem 1. (1 point)
It can be helpful to classify a differential equation, so that we can
predict the techniques that might help us to find a function which
solves the equation. Two classifications are the order of the equa-
tion – (what is the highest number of derivatives involved) and
whether or not the equation is linear .
Linearity is important because the structure of the the family of
solutions to a linear equation is fairly simple. Linear equations
can usually be solved completely and explicitly.
Determine whether or not each equation is linear:
? 1. t2d2y
dt2+tdy
dt +2y=sint
? 2. d3y
dt3+tdy
dt + (cos2(t))y=t3
? 3. d4y
dt4+d3y
dt3+d2y
dt2+dy
dt =1
? 4. y00 −y+t2=0
Solution:
SOLUTION
1. The DE is 2nd order since the highest derivative is second
order. The DE is linear since it has the form a2(t)d2y
dt2+a1(t)dy
dt +
a0(t)y=g(t).
2. The DE is 3rd order since the highest derivative is third or-
der. The DE is linear since it has the form a3(t)d3y
dt3+a2(t)d2y
dt2+
a1(t)dy
dt +a0(t)y=g(t).
3. The DE is 4th order since the highest derivative is fourth or-
der. The DE is linear since it has the form a4(t)d4y
dt4+a3(t)d3y
dt3+
a2(t)d2y
dt2+a1(t)dy
dt +a0(t)y=g(t).
4. The DE is 2nd order since the highest derivative is sec-
ond order. The DE is Linear since it can be put in the form
a2(t)d2y
dt2+a1(t)dy
dt +a0(t)y=g(t).
Correct Answers:
•2LINEAR
•3LINEAR
•4LINEAR
•2LINEAR
Problem 2. (1 point)
Determine the order of the following differential equations and
whether they are linear or non linear.
? 1. d3y
dt3+tdy
dt + (cos2(t))y=t3
? 2. y00 −y+t2=0
? 3. d2y
dt2+sin(t+y) = sint
? 4. dy
dt +ty2=0
Correct Answers:
•3LINEAR
•2LINEAR
•2NONLINEAR
•1NONLINEAR
Problem 3. (1 point)
Which of the following are first order linear differential equations?
•A. d2y
dx2+sin(x)dy
dx =cos(x)
•B. dy
dx =y2−3y
•C. sin(x)dy
dx −3y=0
•D. dy
dx 2+cos(x)y=5
•E. dP
dt +2tP =P+4t−2
•F. xdy
dx −4y=x6ex
Correct Answers:
•CEF
1
Problem 4. (1 point)
In problems below, (a) identify the independent variable and the
dependent variable of each equation (use ’t’ for the independent
variable if an independent variable is not given explicitly); (b)
give the order of each differential equation (enter ’1’ for first or-
der, ’2’ for second order and so on; do not include the quotes);
and (c) state whether the equation is linear or nonlinear. If your
answer to (c) is nonlinear, make sure that you can explain why
this is true.
equation (a) independent (a) dependent (b) order (c) linear/nonlinear
y0=y−x2[?/linear/nonlinear]
xy0=2y[?/linear/nonlinear]
x00 +5x=e−x[?/linear/nonlinear]
Correct Answers:
•x
•y
•1
•linear
•x
•y
•1
•linear
•t
•x
•2
•nonlinear
Problem 5. (1 point)
Which of the following functions are solutions of the differential
equation y00 −2y0−15y=0?
•A. y(x) = ex
•B. y(x) = e−x
•C. y(x) = 5x
•D. y(x) = −3x
•E. y(x) = e5x
•F. y(x) = 0
•G. y(x) = e−3x
Correct Answers:
•EFG
Problem 6. (1 point)
Match each of the following differential equations with a solution
from the list below.
1. y00 +12y0+36y=0
2. y00 −12y0+36y=0
3. y00 +y=0
4. 2x2y00 +3xy0=y
A. y=e−6x
B. y=cos(x)
C. y=e6x
D. y=1
x
Correct Answers:
•A
•C
•B
•D
Problem 7. (1 point)
Find the value of kfor which the constant function x(t) = kis a
solution of the differential equation 2t4dx
dt +5x−3=0.
Solution:
SOLUTION
Since the function x(t) = kis constant, we have dx
dt =0. Substitut-
ing into the differential equation yields
5k−3=0.
Thus x(t) = kis a solution of the differential equation if
k=3
5
Correct Answers:
•0.6
Problem 8. (1 point)
For what values of rdoes the function y=9erx satisfy the differ-
ential equation y00 +y0−20y=0?
The smaller one is .
The larger one (possibly the same) is .
Correct Answers:
•-5
•4
2
Problem 9. (1 point)
Find ksuch that x(t) = 5tis a solution of the differential equation
dx
dt =kx.
k=.
Solution:
SOLUTION
Substituting x=5tand dx
dt =5tln(5)in the differential equation,
yields
5tln(5) = k5t.
Thus x(t)is a solution of the differential equation if
k=ln(5)
Correct Answers:
•1.6094379124341
Problem 10. (1 point)
For what positive values of kdoes the function y=sin(kt)satisfy
the differential equation y00 +36y=0?
For what negative values of kdoes the function y=cos(kt)satisfy
the differential equation y00 +36y=0?
Correct Answers:
•6
•-6
Problem 11. (1 point)
It is easy to check that for any value of c, the function
y(x) = ce−2x+e−x
is solution of equation
y0+2y=e−x.
Find the value of cfor which the solution satisfies the initial con-
dition y(3) = 5.
c=
Solution:
SOLUTION
Substituting the initial condition gives
y(3) = ce−6+e−3=5
Thus
c= (5−e−3)e6=5e6−e3
Correct Answers:
•1997.05843054049
Problem 12. (1 point)
It is easy to check that for any value of c, the function
y=x2+c
x2
is solution of equation
xy0+2y=4x2,(x>0).
Find the value of cfor which the solution satisfies the initial con-
dition y(1) = 10.
c=
Solution:
SOLUTION
Substituting the initial condition gives
y(1) = 1+c
1=10
Thus
c=1(10 −1) = 9
Correct Answers:
•9
Problem 13. (1 point)
The solution of a certain differential equation is of the form
y(t) = aexp(6t) + bexp(8t),
where aand bare constants.
The solution has initial conditions y(0) = 4 and y0(0) = 3.
Find the solution by using the initial conditions to get linear equa-
tions for aand b.
y(t) =
Solution:
SOLUTION
We have y0=6ae6t+8be8t. Substituting the initial conditions
yields the system:
4=a+b
3=6a+8b
Solving the system yields a=29
2and b=−21
2.
Thus the solution with the given initial conditions is
y=29
2e6t−21
2e8t
Correct Answers:
•14.5 *(eˆ(6 *t)) + -10.5 *(eˆ(8 *t))
3
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