The general solution of the differential equa | vedclass

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(C) Given the differential equation: $(x^3-3xy^2)dx = (y^3-3x^2y)dy$ $\Rightarrow \frac{dy}{dx} = \frac{x^3-3xy^2}{y^3-3x^2y}$ This is a homogeneous d

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$c^2(x^2+y^2)^2 = (y^2-x^2)$

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(C) Given the differential equation: $(x^3-3xy^2)dx = (y^3-3x^2y)dy$ $\Rightarrow \frac{dy}{dx} = \frac{x^3-3xy^2}{y^3-3x^2y}$ This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting into the equation: $v + x\frac{dv}{dx} = \frac{x^3-3x(vx)^2}{(vx)^3-3x^2(vx)} = \frac{x^3(1-3v^2)}{x^3(v^3-3v)} = \frac{1-3v^2}{v^3-3v}$ $x\frac{dv}{dx} = \frac{1-3v^2}{v^3-3v} - v = \frac{1-3v^2-v^4+3v^2}{v^3-3v} = \frac{1-v^4}{v^3-3v}$ Separating variables: $\int \frac{v^3-3v}{1-v^4} dv = \int \frac{dx}{x}$ $\int \frac{v^3}{1-v^4} dv - 3\int \frac{v}{1-v^4} dv = \ln|x| + C$ Let $1-v^4 = t \Rightarrow -4v^3 dv = dt \Rightarrow v^3 dv = -\frac{dt}{4}$. Let $v^2 = m \Rightarrow 2v dv = dm \Rightarrow v dv = \frac{dm}{2}$. $-\frac{1}{4}\ln|1-v^4| - \frac{3}{2}\int \frac{dm}{1-m^2} = \ln|x| + C$ $-\frac{1}{4}\ln|1-v^4| - \frac{3}{4}\ln|\frac{1+v^2}{1-v^2}| = \ln|x| + C$ Substituting $v = \frac{y}{x}$: $-\frac{1}{4}\ln|1-\frac{y^4}{x^4}| - \frac{3}{4}\ln|\frac{1+y^2/x^2}{1-y^2/x^2}| = \ln|x| + C$ Simplifying leads to the solution: $c^2(x^2+y^2)^2 = (y^2-x^2)$.

The general solution of the differential equation $(x^3-3xy^2)dx = (y^3-3x^2y)dy$ is,where $c$ is an arbitrary constant:

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