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 $.
$ 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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