Exact Differential Equations

  • #1
I thought I got this problem wrong, but I think I have it right now. It turned out that when I was taking the derivative of e^xy2with respect to y, I forgot that you're supposed to multiply by 2xy (the derivative of xy2), not just x.

Homework Statement



(y2* e^xy2 + 4x3 dx + (2xy * e^xy2 - 3y2 dy = 0

Homework Equations



The Attempt at a Solution



First, I test to make sure it's exact. When I differentiate the first with respect to y, and the second with respect to x, I get 2xy3 * e^xy2 + 2y*e^xy2 for both, so it is indeed exact.

When I integrate (y2* e^xy2 + 4x3 dx, I get e^xy2 + x^4 + f(y)

The derivative of e^xy2 + x^4 + f(y) with respect to y is 2xy*e^xy2 + f'(y), and f'(y) is (2xy * e^xy2 - 3y2 dy), so I integrate 3y2 with respect to y, and get y3

So my final function is f(x,y) = e^xy2 + x4 + y3
 
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Answers and Replies

  • #2
ehild
Homework Helper
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I thought I got this problem wrong, but I think I have it right now. It turned out that when I was taking the derivative of e^xy2with respect to y, I forgot that you're supposed to multiply by 2xy (the derivative of xy2), not just x.

Homework Statement



(y2* e^xy2 + 4x3 dx + (2xy * e^xy2 - 3y2 dy = 0

I think you meant the equation (y2* e^(xy2) + 4x3)dx + (2xy * e^(xy2) - 3y2 )dy = 0. Do not forget parentheses!

First, I test to make sure it's exact. When I differentiate the first with respect to y, and the second with respect to x, I get 2xy3 * e^xy2 + 2y*e^xy2 for both, so it is indeed exact.

When I integrate (y2* e^xy2 + 4x3 dx, I get e^xy2 + x^4 + f(y)

The derivative of e^xy2 + x^4 + f(y) with respect to y is 2xy*e^xy2 + f'(y), and f'(y) is (2xy * e^xy2 - 3y2 dy), so I integrate 3y2 with respect to y, and get y3

So my final function is f(x,y) = e^xy2 + x4 + y3

Excellent! But you need to include the integration constant, so f(x,y) = e^(xy2) + x4 + y3 + C

ehild
 

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