Query 06

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course Mth 279

April 3 around 7pm.

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:

M(t,y) = 3t^2y

N(t,y) = 6t + y^3

M_y = 3t^2

N_t = 6

M_y does NOT equal N_t, therefore 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:

M(t,y) = [6y + 9/2t^2y^2 + t]

N(t,y) = [6t + 3t^3]

M_y = 9yt^2 + 6

N_t = 9t^2 + 6

M_y does NOT equal N_t, therefore, it’s not exact.

confidence rating #$&*:

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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:

0 = [( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)] - y’

0 = [( y cos(t y) + 1)] - [(t cos(t y) + 2 y e^y^2)y’]

M(t,y) = (y cos (ty) + 1)

N(t,y) = (t cos(t y) + 2 y e^y^2)

M_y = cos (ty) - y*sin(ty)*t

N_t = cos (ty) - t*sin(ty)*y

EXACT.

Y(0) = pi

H_t = M(t,y)

H_y = N(t,y)

H(t,y) = C

Int (y cos (ty) + 1 dt) = H(t,y)

H(t,y) = sin(ty) + t + g(t)

H_t = cos(ty) * y + 1 + (dg/dt)

(dg/dt) = C1 = g(t)

This is where I got lost….

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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:

If this is exact, that means that M_y must equal N_t.

M_y = [2y*sin(t)]

N_t has to equal the same thing.

confidence rating #$&*:

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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:

M(t,y) = (ay + bt)

N(t,y) = (y + at)

M_y = a

N_t = a

If y = -t - sqrt( 4 - t^2 ) is the solution, you will have to undo the implicit differentiation, and see what H(t,y) and g(t) is and you will then find out what the values of the following variables, knowing that H_t = M(t,y) and H_y = N(t,y).

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: