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Sunday, December 27, 2015

Exact differential equation

Problem:

$ 3x^2y dx + x^3dy = 0 $

Solution:

We have $ \frac{\partial 3x^2y}{\partial y} = 3x^2 = \frac{\partial x^3}{\partial x} $, therefore, we can find $ f(x, y) $ such that $ \frac{\partial f}{\partial x} = 3x^2y $ and $ \frac{\partial f}{\partial y} = x^3 $

It is quite obvious that $ f(x, y) = x^3y + c $.

The answer to the differential equation is therefore $ y = cx^{-3} $.

As a quick check, we have $ \frac{dy}{dx} = -3cx^{-4} $, therefore

$ 3x^2y dx + x^3dy = 3x^2(cx^{-3})dx + x^3(-3cx^{-4})dx = 0 $.

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