Question

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

• A. $x^2(2xy+y^2)=c^2$
• B. $x^2(2xy-y^2)=c^2$
• C. $x^2(y^2-2x{y}^2)=c^2$
• D. none of these

Answer 1

The correct option is A

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

Explanation for the correct option

Given: $\left(3xy+y^2\right)dx+\left(x^2+xy\right)dy=0$

$$\begin{gathered} \Rightarrow \left(3xy+y^2\right)dx=-\left(x^2+xy\right)dy \\ \Rightarrow \frac{dy}{dx}=\frac{-\left(3xy+y^2\right)}{\left(x^2+xy\right)} \end{gathered}$$

Let $y=vx$

$\frac{dy}{dx}=x\frac{dv}{dx}+v$

therefore

$$\begin{gathered} \Rightarrow x\frac{dv}{dx}+v=\frac{-(3x^{2}v+v^2{x}^2)}{(x^2+x^{2}v)} \\ \Rightarrow x\frac{dv}{dx}=\frac{-v-v^2-3v-v^2}{(1+v)} \\ \Rightarrow x\frac{dv}{dx}=\frac{-2v^2-4v}{(1+v)} \\ \Rightarrow -2\frac{dx}{x}=\frac{(1+v)dv}{\left(v+2\right)v} \end{gathered}$$

Integrating both sides

$$\begin{gathered} \Rightarrow -2\log \left(x\right)=\int \left(\frac{1}{2\left(2+v\right)}+\frac{1}{2v}\right)dv \\ \Rightarrow -4\log \left(x\right)=\int \left(\frac{1}{\left(2+v\right)}\right)dv+\int \left(\frac{1}{v}\right)dv \\ \Rightarrow -4\log \left(x\right)=\log (2+v)+\log \left(v\right)+\log \left(c\right) \\ \Rightarrow \log \left(x^4\right)+\log (2+v)+\log \left(v\right)=\log \left(c\right) \\ \Rightarrow \log (x^4·(2+v\left)\right(v\left)\right)=\log \left(c\right) \\ \Rightarrow \log \left[x^2·\left(2x+y\right)\left(y\right)\right]=\log \left(c\right) \end{gathered}$$

taking antilog both sides

$x^2·\left(2xy+y^2\right)=c$

Hence, option (A) is correct.