Word Count : 632

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We are asked to solve the differential equation: $$ y' = \frac{zy^4 + x^4}{x y^3}. $$ First, let's clarify the notation. Assuming $z = y'$, the equation becomes: $$ y' = \frac{y' y^4 + x^4}{x y^3}. $$ Multiply both sides by $x y^3$: $$ x y^3 y' = y' y^4 + x^4. $$ Bring terms containing $y'$ to one side: $$ x y^3 y' - y^4 y' = x^4. $$ Factor $y'$ on the left: $$ y'(x y^3 - y^4) = x^4. $$ So we get: $$ y' = \frac{x^4}{x y^3 - y^4} = \frac{x^4}{y^3 (x - y)}. $$ Now we have a separable form: $$ \frac{dy}{dx} = \frac{x^4}{y^3 (x - y)}. $$ Let us try the substitution $v = \frac{y}{x} \implies y = vx \implies y' = v + x \frac{dv}{dx}$. Substitute into the equation: $$ v + x \frac{dv}{dx} = \frac{x^4}{(vx)^3 (x - vx)} = \frac{x^4}{v^3 x^3 \cdot x(1 - v)} = \frac{1}{v^3 (1-v)}. $$ Rewriting: ______ ______ ___ ______ ____ _________ ______.
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