The general solution of the differential equa | vedclass

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(C) Given differential equation is $x^2 y dx - (x^3 + y^3) dy = 0$.Rearranging the terms,we get $\frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3}$.This is a ho

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$\log |y| - \frac{x^3}{3y^3} = c$

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(C) Given differential equation is $x^2 y dx - (x^3 + y^3) dy = 0$.Rearranging the terms,we get $\frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3} = \frac{vx^3}{x^3(1 + v^3)} = \frac{v}{1 + v^3}$.Then,$x \frac{dv}{dx} = \frac{v}{1 + v^3} - v = \frac{v - v - v^4}{1 + v^3} = \frac{-v^4}{1 + v^3}$.Separating the variables: $\frac{1 + v^3}{v^4} dv = -\frac{dx}{x}$.Integrating both sides: $\int (v^{-4} + v^{-1}) dv = -\int \frac{1}{x} dx$.This gives $-\frac{1}{3v^3} + \log |v| = -\log |x| + c$.Substituting $v = \frac{y}{x}$: $-\frac{1}{3(y/x)^3} + \log |\frac{y}{x}| = -\log |x| + c$.$-\frac{x^3}{3y^3} + \log |y| - \log |x| = -\log |x| + c$.Thus,$\log |y| - \frac{x^3}{3y^3} = c$.

The general solution of the differential equation $x^2 y dx - (x^3 + y^3) dy = 0$ is

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