Initial Value problem

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Ponomous

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Jan 23, 2011
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Find the solution of the initial value problem.
(2x sin y + 3x[sup:1ds0wgl0]2[/sup:1ds0wgl0]y)dx + (x[sup:1ds0wgl0]3[/sup:1ds0wgl0] + x[sup:1ds0wgl0]2[/sup:1ds0wgl0] cos y + y[sup:1ds0wgl0]2[/sup:1ds0wgl0])dy = 0; y(1) = pi/2:

The first thing I tried to do was find the integrating factor but I could not get the equation in the form y' + ay = z so that I could use u = e[sup:1ds0wgl0]int(a dt)[/sup:1ds0wgl0] to find the integrating factor u.

This one is more complicated than the problems I have done in the past, could someone point me in the right direction on where to start?

Thanks
 
I'm hoping the "y2" should be "y-squared". This would lead to an "Exact" solution.
 
Yes that was supposed to be a y squared, that was my mistake. How do you know that this leads to an exact solution? I know that I am supposed to pick an equation so that the partial x and partial y are equal.

when I go through this I get:

Partial with respect to y of 2x sin y + 3x[sup:1hl4votw]2[/sup:1hl4votw]y = 2x cos y + 3x[sup:1hl4votw]2[/sup:1hl4votw]
and
Partial with respect to x of x[sup:1hl4votw]3[/sup:1hl4votw] + x[sup:1hl4votw]2[/sup:1hl4votw]cosy + y[sup:1hl4votw]2[/sup:1hl4votw] = 3x[sup:1hl4votw]2[/sup:1hl4votw] +2xcosy
Which are equal.

So, now that I have this, I am lost. What is the next step?
 
Those are the partial derivatives. Now you need an anti-derivative ("partial"?). Pick the easier one. I'll use the first.

\(\displaystyle \int (2x\cdot sin(y) + 3x^{2}y)\;dx\;=\;x^{2}\sin(y) + x^{3}y + g(y)\)

g(y) is some, as yet unknown, function of y.

This, then, is the generator for both pieces with which you started. Its partial derivatives should be those things with which you started. We know already that the 'dx' piece works (because we made it work). It's time for the 'dy' piece.

\(\displaystyle \frac{d}{dy}\;\left [x^{2}\sin(y) + x^{3}y + g(y)\right ]\;=\;x^{2}\cos(y) + x^{3} + g'(y)\;=\;x^{3} + x^{2} \cos(y) + y^{2}\)

After a little algebra and another brief anti-derivative, you're nearly done.
 
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