Assignment 4

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course Mth 279

6/20 1:30

Solve each equation:*********************************************

Question: 1. y ' + y = 3

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Your solution:

p(t) = 1

int(p(t)) = int(1)dt = t

general form: y = Ce^(-int(p(t))) + e^(-int(p(t)))*int(e^(int(p(t))) * g(t))

y = Ce^-t + e^-t * int(e^t * 3 dt

y = Ce^-t + 3e^-t * int(e^t)dt

y = Ce^-t + 3e^-t *e^t

y = Ce^-t + 3

confidence rating #$&*:

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1

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Given Solution:

???Are you going to provide the given solutions to the homework’s once they are submitted???

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Self-critique (if necessary):

Had to look at chegg to review a similar problem from the book

Was confused but slowly worked my way through problem

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Self-critique rating:

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Question:

2. y ' + t y = 3 t

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Your solution:

p(t) = t

int(p(t)) = t^2/2

y = Ce^(-t^2/2) + e^(-t^2/2) * int(e^(t^2/2)) *3t dt

do u sub for integral let u= t^2 du= 2t

y = Ce^(-t^2/2) + e^(-t^2/2) * 3/2int(e^.5u) du

y = Ce^(-t^2/2) + e^(-t^2/2) * (3/2)*2e^(t^2/2)

simplifying expression to:

y = Ce^(-t^2/2) + 3

confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

slowly understanding process and general form

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Self-critique rating:

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Question:

3. y ' - 4 y = sin(2 t)

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Your solution:

p(t) = -4 therefore int(p(t)) = -4t

y = Ce^(4t) + e^(4t)*int(e^(-4t)*sin(2t))

use integration by parts twice

u = sin(2t) du = 2cos(2t) dv=e^(-4t) v= (-1/4)e^(-4t)

uv - int(v du)

sin(2t)(-1/4)e^(-4t) - int((-1/4)e^(-4t)*2cos(2t) dt

sin(2t)(-1/4)e^(-4t) + .5int(e^(-4t)*cos(2t))dt

u=cos(2t) du=-2sin(2t) dv=e^(-4t) v=(-1/4)e^(-4t)

sin(2t)(-1/4)e^(-4t) + .5((cos(2t))((-1/4)e^(-4t)) - int((-1/4)e^(-4t)*-2sin(2t)))dt

therefore:

int(e^(-4t)*sin(2t)) = -.25e^(-4t)sin(2t) + (1/8)e^(-4t)cos(2t) - .25int(e^(-4t)*sin(2t))

adding .25int(e^(-4t)*sin(2t)) to both sides you get:

(5/4)int(e^(-4t)*sin(2t)) = -.25e^(-4t)sin(2t) + (1/8)e^(-4t)cos(2t)

= 4/5(-.25e^(-4t)sin(2t) + (1/8)e^(-4t)cos(2t))

= (-1/5)e^(-4t)*sin(2t) - (1/10)e^(-4t)*cos(2t)

Therefore:

y = Ce^(4t) + e^(4t)*((-1/5)e^(-4t)*sin(2t) - (1/10)e^(-4t)*cos(2t))

distributing e^(4t) through

y = Ce^(4t) - (1/5)sin(2t) - (1/10)cos(2t)

confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

I found the integral long and difficult. I did int(e^(-4t)sin(2t) by parts twice; and it could just keep going around in circles because of the oscillating sin and e integrals.

I had to look at wolfram alpha to help with the integral and to see to stop after the second by parts.

Is there another way to go about solving this problem/integral???

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Self-critique rating:

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Question:

4. y ' + y = e^t, y (0) = 2

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Your solution:

p(t) = 1 int(p(t)) = t

y = Ce^-t + e^-t * int(e^t * e^t) dt

y = Ce^-t + e^-t * int(e^(2t)) dt

int(e^(2t)) dt = (1/2)e^(2t)

y = Ce^-t + e^-t*(1/2)e^(2t)

y = Ce^-t + (1/2)e^t

2 = Ce^(-0) + (1/2)e^0

2 = C + (1/2)

C = 3/2

Therefore y = (3/2)e^-t + (1/2)e^t

confidence rating #$&*:

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2

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2

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Your solution:

p(t) = 3 int(p(t))= 3t

y = Ce^(-3t) + e^(-3t)*int(e^(3t)*(3+2t+e^t))dt

int(e^(3t)*(3+2t+e^t))dt

distribute e^(3t) through and then make three separate integrals

int(3e^(3t))dt + 2int(te^(3t))dt + int(e^(4t))dt

middle integral done by parts where u = t du= dt dv=e^(3t) v=(1/3)e^(3t)

3((1/3)e^(3t) + (1/3)te^(3t) - (1/9)e^(3t) + (1/4)e^(4t)

y = Ce^(-3t) + e^(-3t)(e^(3t) + (1/3)te^(3t) - (1/9)e^(3t) + (1/4)e^(4t))

y = Ce^(-3t) + (8/9) + (1/3)t + (1/4)e^t

e^2 = Ce^(-3) + (8/9) + (1/3) + (1/4)e^1

Ce^(-3) = e^2 - (11/9) - (1/4)e

C = e^5 - (11/9)e^3 - (1/4)e^4

y = (e^5 - (11/9)e^3 - (1/4)e^4)*e^(-3t) + (8/9) + (1/3)t + (1/4)e^t

@&

The correct solution would be

y(t) = (e^5 - (1/4)e^4 - (13/9)e^3) e^(-3t) + (2/3)t + (1/4)e^t + (7/9)

You have much of this solution so you're definitely on the right track here.

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confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

I am not 100% confident on the math. I feel like I might have made a mathematical error

@&

It appears that you did, but your overall procedure was sound.

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Self-critique rating:

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Question:

6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)?

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Your solution:

y = Ce^(-t^2)+1 t>0

general y = Ce^(-int(p(t)) + e^(-int(p(t)))*int(e^(int(p(t))) * g(t)

-int(p(t)) = -t^2 take derivative to find p(t)

-p(t) = - 2t

p(t) = 2t

y = Ce^(-t^2) + e^(-t^2)*int(e^(t^2))*g(t) dt

We know that:

e^(-t^2)*int(e^(t^2))*g(t) = 1

Take derivative of both sides

int(e^(t^2)g(t))dt = e^(t^2)

e^(t^2)g(t) = 2te^(t^2)

g(t) = (2t*e^(t^2))/(e^(t^2))

g(t) = 2t

confidence rating #$&*:

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1.5

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

Self-critique (if necessary):

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Self-critique rating:

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@&

I agree with most of your solutions.

Check my notes and let me know if you have questions. You can always use the question form if you do.

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