What is the solution for 4x^2ydx+xydy=0 ?

The solution for 4x^2ydx+xydy=0 is y=-2x^2+c_{1}

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4x^2ydx+xydy=0

4 x^{2} y d x + x y d y = 0

y = - 2 x^{2} + c_{1}

Solution steps

4 x^{2} y d x + x y d y = 0

Let

y

be the dependent variable. Divide by

d x

4 x^{2} y + x y \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

4 x^{2} y + x y y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = - 4 x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 4 x d x

\int - 4 x d x = - 2 x^{2} + c_{1}

y = - 2 x^{2} + c_{1}

\int_{0}^{5} t e^{- t} d t

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\int \frac{1}{sec ^{2} ( x )} d x

\int_{0}^{1} sin (2 π x) d x

What is the solution for 4x^2ydx+xydy=0 ?
The solution for 4x^2ydx+xydy=0 is y=-2x^2+c_{1}