Alyssa Kritz Elzanowski MAT 275 Spring 2020
Assignment Section 2.1 Integrating Factor due 01/27/2020 at 11:59pm MST
1. (1 point) Solve the initial value problem:
dy
dx +2y=6,y(0) = 0
y(x) = .
Correct Answers:
•(6/2)*(1-eˆ(- 2*x))
2. (1 point) Solve the initial value problem
dy
dx +ycosx=8cosx,y(0) = 10
y(x)= .
Correct Answers:
•8 + 2*eˆ(-sin(x))
3. (1 point)
Find the general solution, y(t), which solves the problem
below, by the method of integrating factors.
9tdy
dt +y=t5,t>0
Put the problem in standard form.
Then find the integrating factor, µ(t) = ,
and finally find y(t) = . (use Cas the
unkown constant.)
Correct Answers:
•tˆ(1/9)
•(tˆ5)/46+C*tˆ(-1/9)
4. (1 point)
Solve the following initial value problem:
tdy
dt +3y=6t
with y(1) = 5.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Correct Answers:
•tˆ(3)
•( 6 * t / (1 +3)) + 3.5 * (t**(-3 ))
5. (1 point)
Solve the initial value problem
5sin(t)dy
dt +cos(t)y=cos(t)sin6(t),
for 0 <t<πand y(π/2) = 15.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Correct Answers:
•sin(t)
•14.9714285714286/sin(t) + 0.0285714285714286 * ((sin(t))**6 )
6. (1 point)
Solve the initial value problem
8(t+1)dy
dt −6y=12t,
for t>−1 with y(0) = 2.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Correct Answers:
•(t+1)ˆ(-0.75)
•6*t +8 + (-6 * ((t + 1)**0.75))
7. (1 point)
Solve the following initial value problem:
dy
dt +0.9ty =4t,y(0) = 8
y(t) = .
Correct Answers:
•(4 / 0.9 ) + 3.55555555555556 * exp( -0.9 * t * t /2)
8. (1 point)
Find the solution of the following IVP:
dy
dt −2ty =−6t2et2,y(0) = −4.
y(t) = .
Correct Answers:
•(-2*tˆ3 + -4)*eˆ(tˆ2)
1
9. (1 point) Find the general solution to the differential equa-
tion
x2+2xy +xdy
dx =0
Put the problem in standard form.
Find the integrating factor, ρ(x) = .
Find y(x) = .
Use Cas the unknown constant.
Correct Answers:
•eˆ(2 x)
•C eˆ(-2 x) - x/2 + 1/2ˆ2
10. (1 point)
Solve the initial value problem
dy
dt −y=3et+15e6t,y(0) = 5
y(t) = .
Correct Answers:
•(5 - 3 )* exp(t) + 3 * t * exp(t) + 3 * exp(6 *t)
11. (1 point) Consider the initial value problem
y0−5y=20t+4et,y(0) = y0
(a) Solve the initial value problem. (enter y0 for y0).
y(t)=
(b) Determine the value of y0that separates solutions that
grow positively as t→∞from those that grow negatively.
y0=
Correct Answers:
•-4*t-4*eˆt/4-4/5+eˆ(5*t)*(y0+9/5)
•-1.8
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