Question

The solutions of $v=u\frac{dv}{du}+{\left(\frac{dv}{du}\right)}^2$ where $u=y$ and $v=xy$ are:

• A. $y=0$
• B. $y=-4x$
• C. $xy=cy+c^2$
• D. $x^{2}y=cy+c^2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $y=0$
B $xy=cy+c^2$
C $y=-4x$

$v=u\frac{dv}{du}+{\left(\frac{dv}{du}\right)}^2$

Differentiating w.r.t $u$, we get
$\frac{dv}{du}=\frac{dv}{du}+u\frac{d^{2}v}{d{u}^2}+2\frac{dv}{du}\frac{d^{2}v}{d{u}^2}$

$\Rightarrow (u+2\frac{dv}{du})\frac{d^{2}v}{d{u}^2}=0$

$\Rightarrow \frac{dv}{du}=c$ or $\frac{dv}{du}=-\frac{u}{2}$

Putting $\frac{dv}{du}=c$, we get

$v=cu+c^2\Rightarrow xy=cy+c^2$

Again putting $\frac{dv}{du}=-\frac{u}{2}$, we get

$v=-\frac{u^2}{4}\Rightarrow y^2=-4xy\Rightarrow y=0$ or $y=-4x$