Question

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

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

Answer 1

Answer: (A)

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