Exact Differential Equations: Solving for f(x,y)

[checkmatechamp]

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

 

[ehild]

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