Question

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

Answer 1

Given, $x\frac{dy}{dx}-2y=x^4{y}^2$
Dividing by $x{y}^2$, we get
$\frac{1}{y^2}\frac{dy}{dx}-\frac{2}{y}\frac{1}{x}=x^3\cdots \left(1\right)$
This is a Bernoulli differential equation.
Put $-\frac{2}{y}=t\Rightarrow \frac{2}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$
Equation $\left(1\right)$ becomes
$\frac{1}{2}\frac{dt}{dx}+\frac{t}{x}=x^3$
$\Rightarrow \frac{dt}{dx}+\frac{2}{x}t=2x^3$
I.F. $=e^{\int \frac{2}{x}dx}=e^{2\ln x}=e^{\ln x^2}=x^2$

So, general solution is given by
$t\cdot x^2=\int (x^2\cdot 2x^3)dx=2\int x^{5}dx$
$\Rightarrow x^{2}t=\frac{2x^6}{6}+C$
$\Rightarrow -\frac{2x^2}{y}=\frac{x^6}{3}+C$
$\Rightarrow -\frac{2}{y}=\frac{x^4}{3}+\frac{C}{x^2}$
If $x=1,y=-6\Rightarrow C=0$
$\therefore$ The particular solution is
$-\frac{2}{y}=\frac{x^4}{3}\Rightarrow y=-\frac{6}{x^4}$
i.e., $f\left(x\right)=-\frac{6}{x^4}$

Now, $f^{\prime }\left(x\right)=24x^{-5}$
$\Rightarrow f^{\prime }\left(3^{1/5}\right)=24\cdot (3{)}^{-5/5}=\frac{24}{3}=8$
Hence, $f^{\prime }\left(3^{1/5}\right)=8$
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