Query 06

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course Mth 279

7/1 9

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) y' + 3*t^2*y = 0

Since in the form Mdt + Ndy = 0

M = 3*t^2*y

N = (6t + y^3)

dM/dy = 3t^2

dN/dt = 6

Because dM/dy does not equal dN/dt the equation is not exact.

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Given Solution:

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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:

(6t + 3t^3 ) y' + 6y + 9/2*t^2*y^2 + t = 0

In the form Mdt + Ndy = 0

M = 6*y + 9/2*t^2*y^2 + t

N = (6*t + 3t^3)

dM/dy = 9yt^2 + 6

dN/dt = 9t^2 + 6

These are not eqivilant, so the euation is not exact.

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Given Solution:

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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:

y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^(y^2))

(t cos(t y) + 2 y e^(y^2))y ' = - ( y cos(t y) + 1)

(t cos(t y) + 2 y e^(y^2))y ' + ( y cos(t y) + 1) = 0

M = t cos(t y) + 2 y e^(y^2)

and

N = y cos(t y) + 1

M_t = cos(t y) - y t sin(t y)

N_y = cos(t y) - y t sin(t y)

These are equivalent so dF = 0

F_y = t cos(t y) + 2y e^(y^2)

F_t = y cos(t y) + 1

F_y = t cos(t y) + 2y e^(y^2)

F = integral( (y cos(t y) + 1) dt) = -sin(t y) + h(y)

h(y) = e^(y^2)

g(t) = 0

F = -sin(t y) + e^(y^2)

dF = 0

F = c

-sin(t y) + e^(y^2) = c

initial condition y(0) = pi

-sin(0) + e^(pi^2) = c

c = e^(pi^2)

solution is -sin(t y) + e^(y^2) = e^(pi^2)

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Given Solution:

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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(t, y) = N(t, y)

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

dH/dt = t^2 + y^2 sin(t)

H(t, y) = (1/3)t^3 + y^2cos(t) + g(y)

dH/dy = 2ycos(t) + dg/dt

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The second term would be dg/dt, not dg/dy.

Note also that since g is a function of y, dg/dt would be zero.

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2ycos(t) + dg/dt = N(t, y)

N(t, y) = 2ycos(t) + dg/dt

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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:

Working backwards

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

(y + a t) y ' + (a y + b t) = 0

Initial condition y(0) = y_0

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

= -sqrt(4)

= -2

C = -2

d/dt (y + t + sqrt( 4 - t^2 ))

d/dt (( 4 - t^2 )^(1/2) + y + t)

(1/2)(-2t)(4-t^2)^(-1/2) + 1

(-t)(4-t^2)^(-1/2) + 1

d/dy (y + t + sqrt( 4 - t^2 )) = 1

M = 1 - t(4-t^2)^(-1/2) + g(y) = 1 - t(4-t^2)^(-1/2) - 2 = - t(4-t^2)^(-1/2) -1

N = 1 + h(t)

h(t) = - t(4-t^2)^(-1/2) -1

(y + a t) y ' + (a y + b t) = 0

1 = (y + a t)

a = (1-y)/t

(a y + b t) = - t(4-t^2)^(-1/2) -1

((1-y)/t)y + bt = - t(4-t^2)^(-1/2) -1

bt = ((y-1)/t)y - t(4-t^2)^(-1/2) -1

b = ((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

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