Question

The general solution of the differential equation $\left(x-y^2\right) d x+y\left(5 x+y^2\right) d y=0$ is :

• A. $\left(y^2+x\right)^4=C\left|\left(y^2+2 x\right)^3\right|$
• B. $\left( y ^2+2 x \right)^4= C \left|\left( y ^2+ x \right)^3\right|$
• C. $\left|\left(y^2+x\right)^3\right|=C\left(2 y^2+x\right)^4$
• D. $\left|\left(y^2+2 x\right)^3\right|=C\left(2 y^2+x\right)^4$

Answer 1

Answer: (A)

$ \left(x-y^2\right) d x+y\left(5 x+y^2\right) d y=0$
$ \frac{d y}{d x}=\frac{y^2-x}{y\left(5 x+y^2\right)} . \text { Let } y^2=v $
$\frac{2 y d y}{d x}=2\left(\frac{y^2-x}{5 x+y^2}\right) $
$ \frac{d v}{d x}=2\left(\frac{v-x}{5 x+v}\right) v=k x$
$ k + x \frac{ dk }{ dx }=2\left(\frac{ kx - x }{5 x + kx }\right) $
$x \frac{ dk }{ dx }=-\frac{\left( k ^2+3 k +2\right)}{ k +5}$
$\int \frac{(5+k)}{(k+1)(k+2)} d k=\int-\frac{d x}{x}$
$ \int\left(\frac{4}{k+1}-\frac{3}{k+2}\right) d k=-\int \frac{d x}{x}$
$ 4 \ln (k+1)-3 \ln (k+2)=-\ln x+\ln c $
$ \frac{(k+1)^4}{(k+2)^3}=-\ln x+\ln c $
$ c\left(y^2+2 x\right)^3=\left(y^2+x\right)^4$