Word Count : 175

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We have $$ \frac{dy}{dx}=\frac{z y^4+x^4}{x y^3}=\frac{z y^4}{x y^3}+\frac{x^4}{x y^3}=\frac{z y}{x}+\frac{x^3}{y^3}. $$ Make the substitution $u=y^4$. Then $du/dx=4y^3\,dy/dx$. Using the differential equation, $$ \frac{du}{dx}=4y^3\Big(\frac{z y}{x}+\frac{x^3}{y^3}\Big)=\frac{4z y^4}{x}+4x^3=\frac{4z u}{x}+4x^3. $$ This is a linear first-order ODE for $u(x)$: ______ ______ ______ _______ __________ _________ ________ ________.
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