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Using z = y/x to transform the given homogeneous differential equation into a differential equation in z and x. By first solving the transformed equation, find the general solution of the original equation, giving y in terms of x.
z = \frac{y}{x} \rightarrow y = xz \rightarrow \frac{dy}{dx} = x\frac{dz}{dx} + z
x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4y^3}{3xy^2}
x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4z^3x^3}{3xz^2x^2}
\dfrac{d}{dx}(xz) = \dfrac{1 + 4z^3}{3z^2}
xz = \displaystyle\int (\dfrac{1 + 4z^3}{3z^2} )
xz = \frac{-1}{3z} + \frac{2}{3}z^2 + c
subbing z = y/x
x = \frac{-x^2}{3y^2} + \frac{2y}{3x} + \frac{cx}{y}
How do I rearrange from y from here? I just can't seem to do it :\
z = \frac{y}{x} \rightarrow y = xz \rightarrow \frac{dy}{dx} = x\frac{dz}{dx} + z
x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4y^3}{3xy^2}
x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4z^3x^3}{3xz^2x^2}
\dfrac{d}{dx}(xz) = \dfrac{1 + 4z^3}{3z^2}
xz = \displaystyle\int (\dfrac{1 + 4z^3}{3z^2} )
xz = \frac{-1}{3z} + \frac{2}{3}z^2 + c
subbing z = y/x
x = \frac{-x^2}{3y^2} + \frac{2y}{3x} + \frac{cx}{y}
How do I rearrange from y from here? I just can't seem to do it :\
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