The general solution of the differential equa | vedclass

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

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

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(D) Given differential equation is $(3xy+y^2) dx + (x^2+xy) dy = 0$. Rearranging,we get $\frac{dy}{dx} = -\frac{3xy+y^2}{x^2+xy}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3v x^2 + v^2 x^2}{x^2 + vx^2} = -\frac{3v+v^2}{1+v}$. $x\frac{dv}{dx} = -\frac{3v+v^2}{1+v} - v = -\frac{3v+v^2+v+v^2}{1+v} = -\frac{2v^2+4v}{1+v}$. Separating variables: $\int \frac{1+v}{2v^2+4v} dv = -\int \frac{1}{x} dx$. Multiply by $2$: $\int \frac{1+v}{v^2+2v} dv = -2\int \frac{1}{x} dx$. Let $u = v^2+2v$,then $du = (2v+2) dv = 2(v+1) dv$. So,$\frac{1}{2} \int \frac{du}{u} = -2\ln|x| + C'$. $\frac{1}{2} \ln|v^2+2v| = -2\ln|x| + C'$. $\ln|v^2+2v| = -4\ln|x| + C''$. $\ln|v^2+2v| + \ln|x^4| = C''$. $\ln|x^4(v^2+2v)| = C''$. $x^4(\frac{y^2}{x^2} + \frac{2y}{x}) = c$. $x^2(y^2+2xy) = c$.

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

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