Assignment 4

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course Mth 279

6/24 10:00

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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2

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Given Solution:

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Self-critique (if necessary):

Good. Danielle, student, watched lectures and taught me after time constraints.

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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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2

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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):

Hard integral. Technique was new.

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

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The way you solved this is the most efficient.

Be sure you understand how multiplying through by e^(-4 t) makes the left-hand side integrable by making it the differential of y e^(-4 t).

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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

@&

You don't appear to show the multiplication by the integrating factor, nor do you express the resulting left-hand side as the derivative of y e^t. This is consistent with not really understanding the method.

However you did get the right final result.

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confidence rating #$&*:

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2

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Given Solution:

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Self-critique (if necessary):

Easy in comparison to number 3

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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) + (2/3)te^(3t) - (2/9)e^(3t) + (1/4)e^(4t))

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

e^2 = Ce^(-3) + (7/9) + (2/3) + (1/4)e^1

Ce^(-3) = e^2 - (13/9) - (1/4)e

C = e^5 - (14/9)e^3 - (1/4)e^4

y = (e^5 - (14/9)e^3 - (1/4)e^4)*e^(-3t) + (7/9) + (2/3)t + (1/4)e^t

confidence rating #$&*:

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0

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Given Solution:

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Self-critique (if necessary):

Unsure if final answer is correct.

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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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2

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Given Solution:

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Self-critique (if necessary):

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&#Your work looks good. See my notes. Let me know if you have any questions. &#