qa 2

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course 279

2-17 1

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

Question: 1. y ' + y = 3

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Your solution:

where p(t) = 1 and equation is y’ + p(t)y = 3

we mult by integrating constant, e^(Int( p(t) dt) = e^(t)

e^(t)y’ + e^(t)y = 3e^(t), we can show (e^(t)y)’ = e^(t)y’ + e^(t)y

(e^(t)y)’ = using product rule

= e^t*y’ + (e^t)’*y

= e^t*y’ + e^t*y, now we put into the form

(e^(t)y)’ = 3e^(t), so taking integral of both sides

e^(t)y = Int( 3e^t dt) = 3Int(e^t dt) = 3e^t + c, we can divide out e^t

y = 3 + Ce^-t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question:

2. y ' + t y = 3 t

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Your solution:

where p(t) = t and equation is y’ + ty = 3t

we mult by integrating constant, e^(Int( t dt) = e^(t^2/2)

e^(t^2/2)y’ + e^(t^2/2)ty = 3t e^(t^2/2), we can show (e^(t^2/2)y)’ = e^(t^2/2)y’ + e^(t^2/2)ty

( e^(t^2/2)y)’ = using product rule

= e^(t^2/2)*y’ + (e^(t^2/2))’*y

(e^(t^2/2))’ is composite fn equal to t* e^(t^2/2))

= e^(t^2/2)*y’ + e^(t^2/2)t*y, now we put into the form

( e^(t^2/2))y)’ = 3t e^(t^2/2), so taking integral of both sides

e^(t^2/2))y = Int( 3t e^(t^2/2)dt) = 3Int(te^(t^2/2))dt) = 3 e^(t^2/2)+ c, we can divide out e^t

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

Confidence rating: 3

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question:

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

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Your solution:

where p(t) = -4 and equation is y’ - 4y = 3t

we mult by integrating constant, e^(Int( -4 dt) = e^(-4t)

e^(-4t)y’ + e^(-4t)ty = sin(2t) e^(-4t), we can show (e^(-4t)y)’ = e^(t^2/2)y’ - 4e^(-4t) y

( e^(-4t)y)’ = using product rule

= e^(t^2/2)*y’ + (e^(-4t))’*y

(e^(-4t))’ is composite fn equal to -4* e^(-4t))

= e^(-4t)*y’ + -4 e^(-4t)t*y, now we put into the form

( e^(-4))y)’ = sin(2t) e^(-4t), so taking integral of both sides

e^(-4t)y = Int(sin(2t) e^(-4t) dt)

Int(sin(2t) e^(-4t) dt) = (1/20)e^(-4t)(-4sin(2t) -2 cos(2t) ) +c

y = (-4sin(2t) -2 cos(2t) )/20 + C/e^(-4t)

= (-4sin(2t) -2 cos(2t) )/20 + Ce^(-4t)

confidence rating #$&*:

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@& Good.*@

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question:

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

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Your solution:

where p(t) = 1 and equation is y’ + p(t)y = e^t

we mult by integrating constant, e^(Int( p(t) dt) = e^(t)

e^(t)y’ + e^(t)y = e^(t)*e^(t), we can show (e^(t)y)’ = e^(t)y’ + e^(t)y

(e^(t)y)’ = using product rule

= e^t*y’ + (e^t)’*y

= e^t*y’ + e^t*y, now we put into the form

(e^(t)y)’ = e^(t)^2 = e^(2t), so taking integral of both sides

e^(t)y = Int(e^2t dt) = (1/2)e^(2t) + c, we can divide out e^2t

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

Now for y(0) = 2, we get

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

=(1/2) + C

C = 1.5, so we end up with..

y = (1/2)e^(t) + 1.5e^-t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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:

where p(t) = 3 and equation is y’ + 3y = 3 + 2 t + e^t

we mult by integrating constant, e^(Int( 3 dt) = e^(3t)

e^(3t)y’ + e^(3t)y = 3e^(3t) + 2 t e^(3t) + e^te^(3t), we can show (e^(3t)y)’ = e^(3t)y’ + e^(3t)y

( e^(3t)y)’ = using product rule

= e^(3t)*y’ + (e^(3t))’*y

(e^(3t))’ is composite fn equal to 3* e^(-4t))

= e^(3t)*y’ + 3 e^(3t)t*y, now we put into the form

( e^(3t))y)’ = 3e^(3t) + 2 t e^(3t) + e^te^(3t), so taking integral of both sides

e^(3t)y = Int(3e^(3t) + 2 t e^(3t) + e^te^(3t) dt)

Int(3e^(3t) + 2 t e^(3t) + e^te^(3t) dt) = Int(3e^(3t) dt) + Int(2te^(3t) dt) + ….

Int(3e^(3t) dt) = e^(3t)

Int(2 t e^(3t) dt), using integration by parts

u = 2t, u’ = 2 dt, v’ = e^(3t), v = (e^(3t))/3

Int(2 t e^(3t) dt) = uv - Int(u’v dt)

=2t*(e^(3t)/3) - Int(2*(e^(3t)/3) dt)

Int(2*(e^(3t)/3) dt) = Int((2/3)e^(3t) dt)

=2/9e^(3t), so final soln is

Int(2t*(e^(3t)/3) dt) = (2te^(3t))/3 - 2/9e^(3t)

=2e^(3t)((3t - 1)/9)

Int(e^te^(3t) dt) =Int(e^(4t) dt), using u sub, u = 4x, du = 4 dt

Int( e^(u)/4 du) = e^(u)/4 = e^(4t)/4, So our final integral for right side is….

Int(3e^(3t) + 2 t e^(3t) + e^te^(3t) dt)

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

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

y = 1 + ((6t - 2)/9) + e^(t)/4 + Ce^(-3t), for y(1) = e^2

e^(2) = 1 + ((6 - 2)/9) + e^(1)/4 + Ce^(-3)

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

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

e^2 - (13/9) - (e/4) = approx. 5.265

C = 5.265/e^(-3) = approx. 105.7512, so our final answer is

y = 1 + ((6t - 2)/9) + e^(t)/4 + 105.75e^(-3t)

confidence rating #$&*:

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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:

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:

p(t) = -2t where

e^(Int(p(t) dt)) = e^(Int(-2t dt))

Int(-2t dt) = -2*Int( t dt) = -2*(t^2/2 + c) = -t^2

confidence rating #$&*:

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@& Your work looks really good here.

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