JEE Main 2023 — Differential Equations Question with Solution

Given:

$\frac{dy}{dx}+\frac{5}{x\left(1+x^5\right)}y=\frac{{\displaystyle {\left(1+x^5\right)}^2}}{x^7}$

This is linear differential equation.

$\mathrm{I}.\mathrm{F}.=e^{\int \frac{5}{x\left(1+x^5\right)}dx}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\int \frac{5x^5}{x^6\left(1+x^5\right)}dx}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{\int \frac{5}{x^6\left(x^{-5}+1\right)}dx}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{-\int \frac{-5}{x^6\left(x^{-5}+1\right)}dx}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{-\int \frac{d\left(x^{-5}+1\right)}{\left(x^{-5}+1\right)}}$

$\Rightarrow \mathrm{I}.\mathrm{F}.=e^{-{\log }_e\left(x^{-5}+1\right)}$

$\Rightarrow \mathrm{I}.\mathrm{F}.={\left(x^{-5}+1\right)}^{-1}=\frac{x^5}{\left(x^5+1\right)}$

So, solution is given by

$y\frac{x^5}{\left(x^5+1\right)}=\int \frac{x^5}{\left(x^5+1\right)}\times \frac{{\left(x^5+1\right)}^2}{x^7}dx$

$\Rightarrow \frac{y{x}^5}{\left(x^5+1\right)}=\int \left(\frac{x^5+1}{x^2}\right)dx$

$\Rightarrow \frac{y{x}^5}{\left(x^5+1\right)}=\int \left(x^3+\frac{1}{x^2}\right)dx$

$\Rightarrow \frac{y{x}^5}{\left(x^5+1\right)}=\frac{x^4}{4}-\frac{1}{x}+C$

Since, $y\left(1\right)=2$, so

$\frac{2}{2}=\frac{1}{4}-1+C\Rightarrow C=\frac{7}{4}$

So,

$\frac{y{x}^5}{\left(x^5+1\right)}=\frac{x^4}{4}-\frac{1}{x}+\frac{7}{4}$

Put $x=2$, then we get

$\frac{y\left(2\right)\times 32}{33}=\frac{16}{4}-\frac{1}{2}+\frac{7}{4}$

$\Rightarrow y\left(2\right)=\frac{21\times 33}{128}=\frac{693}{128}$