Question

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

Answer 1

Given, $\frac{x d y}{d x}-2 y=x^4 y^2$ or $\frac{d y}{d x}-\frac{2}{x} y=x^3 y^2$. Now dividing by $y^2$, we get
$\frac{1}{y^2} \frac{d y}{d x}-\frac{2}{y} \frac{1}{x}=x^3$ ....(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
$\Rightarrow \frac{1}{2} \frac{ dt }{ dx }+\frac{ t }{ x }= x ^3 \Rightarrow \frac{ dt }{ dx }+\frac{2}{ x } t =2 x ^3$
$IF = e ^{\int \frac{2}{ x } dx }= e ^{2 \ln x }= e ^{\ln x ^2}= x ^2$
So, general solution is given by
$x^2 t=\frac{2 x^6}{6}+C \Rightarrow \frac{-x^2 \cdot 2}{y}=\frac{x^6}{3}+C \Rightarrow \frac{-2}{y}=\frac{x^4}{3}+\frac{C}{x^2}$
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, $\frac{ dy }{ dx }=24 x ^{-5}=\frac{24}{ x ^5}$. Hence, $\left.\frac{ dy }{ dx }\right]_{ x =3^{\frac{1}{5}}}=\frac{24}{3}=8$.