Query 6

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course MTH 279

2/9 2 pm

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:

Equation is not exact. M_y = 3t^2 and N_t = 6

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

Not an exact equatiom. M_y = 6 + 9t^2y and N_t = 6 + 9t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

F = sin(yt) + t + e^y^2 = e^pi^2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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 = partial derivative wrt y of the parital integral of M wrt t.

N = -2ycos(t) + g'(y)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

y_0 = -1

I didn't know how to find a and b. I got as far as M = -1 +t/(sqrt( 4-t^2) = ay + bt and N = -1 = y + at

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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?

This equation is exact, with M = (y + a t) so that M_y = a, and N = (a y + b t) so that N_t = a.

Thus the equation is of the form dF = 0 with

F = integral( (y + at) dy) = y^2 / 2 + a t y + g(t)

and

F = integral( (a y + b t) dt) = a y t + b t^2 / 2 + h(y).

We reconcile the two forms of F by letting g(t) = b t^2 / 2 and h(y) = y^2 / 2, so that

F = y^2 / 2 + b t^2 / 2 + a t y.

dF = 0 has solution F = c, where c is constant, so the implicit solution is

y^2 / 2 + b t^2 / 2 + a t y = c.

If y = -t - sqrt( 4 - t^2 ) then y^2 = t^2 + (4 - t^2) + 2 t sqrt(4 - t^2) = 4 + 2 t sqr(4 - t^2). Substituting for y in the equation we therefore obtain

4 + 2 t sqr(4 - t^2) + b t^2 / 2 + a t ( -t - sqrt( 4 - t^2) ) = c

(2 t - a t) sqrt( 4 - t^2) + (b/2 - a) t^2 + 4 = c

Thus

2 t - a t = 0

b/2 - a = 0

c = 4

and it follows that

a = 2, b = 4

and when t = 0 we have

y_0^2 / 2 = 4

so that y_0 = 2 sqrt(2).

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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On some problems, especially those on which you were unable to complete the solution, I need to see what you are thinking, so more detail would be very helpful.

In any case, your work looks good overall.

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