Question

A solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}+y=0$ is

• A. $y=2$
• B. $y=2x$
• C. $y=2x-4$
• D. $y=2x^2-4$

Answer 1

The correct option is C $y=2x-4$

Let $p=\frac{dy}{dx}$.

Therefore, the given equation becomes

$p^2-xp+y=0⟹y=px-p^2$

On differentiating wrt x, we get

$\frac{dy}{dx}=p+x\frac{dp}{dx}-2p\frac{dp}{dx}$

$\Rightarrow p=p+x\frac{dp}{dx}-2p\frac{dp}{dx}$

$\Rightarrow \frac{dp}{dx}(x-2p)=0$

$\Rightarrow \frac{dp}{dx}=0$ or $x-2p=0$

Now, $x-2p=0$

$\Rightarrow \frac{dy}{dx}=\frac{x}{2}$

$\Rightarrow y=\frac{x^2}{4}+c$

Also from the other equality,

$\frac{dp}{dx}=0\Rightarrow p=k$ (constant)

Substituting $p=k$ in the given equation, we get

$y=kx-k^2$

On comparing the given options with the obtained solutions

$y=kx-k^2$ and $y=\frac{x^2}{4}+c$

We get that the only possible solution is $y=2x-4$.