The solution of the differential equation (3x | vedclass

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(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$.This can be written as $\frac{dy}{dx} = - \frac{3xy + y^2}{x^2 + xy}$.This i

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

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(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$.This can be written as $\frac{dy}{dx} = - \frac{3xy + y^2}{x^2 + xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = - \frac{3x^2v + x^2v^2}{x^2 + x^2v} = - \frac{v^2 + 3v}{v + 1}$.$x\frac{dv}{dx} = - \frac{v^2 + 3v}{v + 1} - v = - \frac{v^2 + 3v + v^2 + v}{v + 1} = - \frac{2v^2 + 4v}{v + 1} = - \frac{2v(v + 2)}{v + 1}$.Separating the variables: $\frac{v + 1}{v(v + 2)} dv = - \frac{2}{x} dx$.Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \implies v + 1 = A(v + 2) + Bv$.For $v = 0$,$A = 1/2$. For $v = -2$,$B = 1/2$.So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = - \int \frac{2}{x} dx$.$\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = - 2 \ln|x| + C$.$\ln|v(v + 2)| = - 4 \ln|x| + C' \implies \ln|v(v + 2)x^4| = C'$.$v(v + 2)x^4 = c^2$. Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2$.$x^2(y^2 + 2xy) = c^2$.

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

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