Question

Given $\frac{dy}{dx}+\frac{y}{x}=y^2.$
Solution is $kx=e^{1/xy}$
If true enter 1 else enter 0

Answer 1

Given differential equation is $\frac{dy}{dx}+\frac{y}{x}=y^2.$

Put $t=\frac{1}{y}$, we get

$\frac{dt}{dx}=-\frac{1}{y^2}\frac{dy}{dx}$

So the given differential equation will transform into $\frac{dt}{dx}-\frac{t}{x}=-1$

The above equation is linear in $t$.

Integration factor is $e^{\int -\frac{1}{x}dx}=\frac{1}{x}$.

We get, $t=-x\ln x+cx$, where $c$ is constant

Now put $t=\frac{1}{y}$, we get

$-\frac{1}{xy}=\ln x+c$ which implies $kx=e^{-\frac{1}{xy}}$

So, the given solution is wrong.