Elizabeth Arena Sharma MAT 275 ONLINE B Summer 2019
Assignment Section 1.3 Classification of Differential Equations due 07/10/2019 at 11:59pm MST
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 equation – (what is the highest number of derivatives in-
volved) 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. d3y
dt3+tdy
dt + (cos2(t))y=t3
? 2. dy
dt +ty2=0
? 3. y00 −y+t2=0
? 4. (1+y2)d2y
dt2+tdy
dt +y=et
Solution:
SOLUTION
1. 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).
2. The DE is 1st order since the highest derivative is first
order. The DE is Non Linear since it contains y2.
3. 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).
4. The DE is 2nd order since the highest derivative is second
order. The DE is Non Linear since it contains y2.
Answer(s) submitted:
•3Linear
•1Nonlinear
•2Linear
•2Nonlinear
(correct)
Correct Answers:
•3LINEAR
•1NONLINEAR
•2LINEAR
•2NONLINEAR
2. (1 point) Determine the order of the following differential
equations and whether they are linear or non linear.
? 1. d4y
dt4+d3y
dt3+d2y
dt2+dy
dt =1
? 2. y00 −y+y2=0
? 3. d2y
dt2+sin(t+y) = sint
? 4. t2d2y
dt2+tdy
dt +2y=sint
Answer(s) submitted:
•4Linear
•2Nonlinear
•2Nonlinear
•2Linear
(correct)
Correct Answers:
•4LINEAR
•2NONLINEAR
•2NONLINEAR
•2LINEAR
3. (1 point) Which of the following are first order linear
differential equations?
•A. dP
dt +2tP =P+4t−2
•B. xdy
dx −4y=x6ex
•C. d2y
dx2+sin(x)dy
dx =cos(x)
•D. sin(x)dy
dx −3y=0
•E. dy
dx 2+cos(x)y=5
•F. dy
dx =y2−3y
Answer(s) submitted:
•(A,B,D)
(correct)
Correct Answers:
•ABD
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 order, ’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.
1
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]
Answer(s) submitted:
•x
•y
•1
•linear
•x
•y
•1
•linear
•t
•x
•2
•nonlinear
(correct)
Correct Answers:
•x
•y
•1
•linear
•x
•y
•1
•linear
•t
•x
•2
•nonlinear
5. (1 point)
Which of the following functions are solutions of the differ-
ential equation y00 −3y0−10y=0?
•A. y(x) = 0
•B. y(x) = −2x
•C. y(x) = e−2x
•D. y(x) = e−x
•E. y(x) = ex
•F. y(x) = e5x
•G. y(x) = 5x
Answer(s) submitted:
•(A,C,F)
(correct)
Correct Answers:
•ACF
6. (1 point)
Match each of the following differential equations with a
solution from the list below.
1. y00 +y=0
2. 2x2y00 +3xy0=y
3. y00 +11y0+30y=0
4. y00 −11y0+30y=0
A. y=cos(x)
B. y=1
x
C. y=e5x
D. y=e−6x
Answer(s) submitted:
•A
•b
•d
•c
(correct)
Correct Answers:
•A
•B
•D
•C
7. (1 point)
Find the value of kfor which the constant function x(t) = k
is a solution of the differential equation 8t2dx
dt +8x−2=0.
Solution: Since the function x(t) = kis constant, we have
dx
dt =0. Substituting into the differential equation yields
8k−2=0.
Thus x(t) = kis a solution of the differential equation if
k=1
4
Answer(s) submitted:
•1/4
(correct)
Correct Answers:
•0.25
8. (1 point)
For what values of rdoes the function y=4erx satisfy the dif-
ferential equation y00 −13y0+30y=0?
The smaller one is .
The larger one (possibly the same) is .
Answer(s) submitted:
•3
•10
(correct)
Correct Answers:
•3
•10
2
9. (1 point)
Find ksuch that x(t) = 15tis a solution of the differential
equation dx
dt =kx.
k=.
Solution: Substituting x=15tand dx
dt =15tln(15)in the dif-
ferential equation, yields
15tln(15) = k15t.
Thus x(t)is a solution of the differential equation if
k=ln(15)
Answer(s) submitted:
•2.708
(correct)
Correct Answers:
•2.70805020110221
10. (1 point)
For what positive values of kdoes the function y=sin(kt)sat-
isfy the differential equation y00 +49y=0?
For what negative values of kdoes the function y=cos(kt)
satisfy the differential equation y00 +49y=0?
Answer(s) submitted:
•7
•-7
(correct)
Correct Answers:
•7
•-7
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
condition y(0) = 9.
c=
Solution: Substituting the initial condition gives
y(0) = ce0+e0=9
Thus
c= (9−e0)e0=9e0−e0
Answer(s) submitted:
•8
(correct)
Correct Answers:
•8
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
condition y(10) = 3.
c=
Solution: Substituting the initial condition gives
y(10) = 100 +c
100 =3
Thus
c=100(3−100) = −9700
Answer(s) submitted:
•-9700
(correct)
Correct Answers:
•-9700
13. (1 point)
The solution of a certain differential equation is of the form
y(t) = aexp(5t) + bexp(6t),
where aand bare constants.
The solution has initial conditions y(0) = 3 and y0(0) = 3.
Find the solution by using the initial conditions to get linear
equations for aand b.
y(t) =
Solution: We have y0=5ae5t+6be6t. Substituting the initial
conditions yields the system:
3=a+b
3=5a+6b
Solving the system yields a=15 and b=−12.
Thus the solution with the given initial conditions is
y=15e5t−12e6t
Answer(s) submitted:
•15eˆ(5t)-12eˆ(6t)
(correct)
Correct Answers:
•15 *(eˆ(5 *t)) + -12 *(eˆ(6 *t))
3
