query_06

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course MTH 279

I have never been more confused in my life as I was on query_06.

Query 06 Differential Equations******Professor, I really don't understand exact equations...I don't understand the examples in the book, or examples on the internet....****PLEASE HELP!****

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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:

Since the equation is in proper form we do not need to change the form. We can see that M = (3 t^2 y) and N = (6t + y^3),

and M_y = 3t^2 and N_t = 6 + y^3. Integrating M, we get (integral of) (3 t^2 y)dy = 3t^2 + h(x).

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This equation is not exact.

M_y is not equal to N_t.

If the equation is exact then it is of the form

dF(t, y) = 0

which expands to

F_y dt + F_t dt = 0,

i.e., to

N dt + M dy = 0

where N = F_y and M = F_t.

In this case we would have

N_t = M_y = F_t y = F_y t.

However it isn't so in this case, so the equation can't be solved as an exact equation.

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confidence rating #$&*:

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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:

???????

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The equation rearranges to

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

This clearly doesn't satisfy the condition for exactness.

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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:

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The equation can be rearranged to

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

so it is of the form M dy + N dt = 0 with

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

and

N = y cos(t y) + 1

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Test the equation for exactness (it will prove to be exact).

Then integrate as required.

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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:

?????

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What is the function M for this equation?

If N_t = M_y, then, what can be said about N?

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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:

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique (if necessary):

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I've given some guidance on most of these problems.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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