Question

A function $y=f(x)$ satisfies $x{f}^{\prime }(x)-2f(x)=x^4{f}^2(x),\forall x>0$ and $f(1)=-6$. Find the value of $f^{\prime }\left(3^{\frac{1}{5}}\right)$.

Answer 1

Answer: 8

Given, $\frac{xdy}{dx}-2y=x^4{y}^2$ or $\frac{dy}{dx}-\frac{2}{x}y=x^3{y}^2$. Now dividing by $y^2$, we get
$\frac{1}{y^2}\frac{dy}{dx}-\frac{2}{y}\frac{1}{x}=x^3$ ....(i)
This is a Bernouli's differential equation, substituting $\frac{-2}{y}=t$, we get $\Rightarrow \frac{2}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$. So, equation (i) becomes
$\Rightarrow \frac{1}{2}\frac{dt}{dx}+\frac{t}{x}=x^3\Rightarrow \frac{dt}{dx}+\frac{2}{x}t=2x^3$
$IF=e^{\int \frac{2}{x}dx}=e^{2\ln x}=e^{\ln x^2}=x^2$
So, general solution is given by
$x^{2}t=\frac{2x^6}{6}+C\Rightarrow \frac{-x^2\cdot 2}{y}=\frac{x^6}{3}+C\Rightarrow \frac{-2}{y}=\frac{x^4}{3}+\frac{C}{x^2}$
If $x=1,y=-6\Rightarrow C=0$
$\therefore \frac{-2}{y}=\frac{x^4}{3}\Rightarrow y=\frac{-6}{x^4}$ i.e. $f(x)=\frac{-6}{x^4}$
Now, $\frac{dy}{dx}=24x^{-5}=\frac{24}{x^5}$. Hence, ${\frac{dy}{dx}]}_{x=3^{\frac{1}{5}}}=\frac{24}{3}=8$.