Assignment 8

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course Mth 279

6/20 9:05

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

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Your solution:

(6t + y^3)dy/dt + 3t^2y = 0

(6t + y^3) dy + (3t^2y)dt = 0

M = 3t^2y dM/dy = 3t^2

N = 6t + 3y^3 dN/dt = 6

dM/dy does not equal dN/dt

equations are NOT exact

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1

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Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *

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Your solution:

(6 t + 3 t^3 )dy/dt + 6 y + 9/2 t^2 y^2 + t = 0

(6 t + 3 t^3 )dy + (6 y + 9/2 t^2 y^2 + t) dt = 0

M = (6y + (9/2)t^2y^2+6) dM/dy = 6 + (9/2)t^2(2y) + 0 = 6 + 9t^2y

N = 6t + 3t^3 dN/dt = 6 + 9t^2

NOT exact

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2

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Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

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Your solution:

(t cos(t y) + 2 y e^y^2)dy + ( y cos(t y) + 1) dt = 0

M = ycos(ty) + 1 dM/dy = cos(ty) - tysin(ty)

N = t*cos(ty) + 2y*e^y^2 dN/dt = cos(ty) - tysin(ty)

YES they are exact

dF/dy = N = tcos(ty)+2ye^y^2

F(t,y) = int(tcos(ty)+2ye^y^2) = -sin(ty) + e^y^2 + g(t)

dF/dt = M = y*cos(ty) + 1

F(t,y) = int(y*cos(ty) + 1) = -sin(ty) + h(y)

Let h(y) = e^y^2

So F(t,y) = -sin(ty) + e^y^2 + C

C = -sin(pi*0) + e^pi^2

C = e^pi^2

e^pi^2 = -sin(ty) + e^y^2

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Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

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Your solution:

N(y, t) dy + (t^2 + y^2 sin(t)) dt = 0

M = t^2 + y^2sin(t)

N = ?

dM/dy = 2ysin(t)

N = int(2ysin(t))dy

N = -2ycos(t) + C

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When integrating with resped to y, any function of t behaves as a constant. So your C should be an arbitrary function of t.

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Given Solution:

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Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

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Your solution:

1. y = -t - sqrt(4 -t^2)

y = -0 - sqrt(4-0^2)

y = -sqrt(4) = -2

2. No real idea where to start to find a and b values

if you plug in t = 0 and y = -2 you can find a value

(-2 - a(0))(-1) + (a(-2) + b(0)) = 0

2 + (-2a) = 0

a = 1?

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You need to start by solving the differential equation in terms of a and b.

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Given Solution:

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