JEE Main 2023 — Mathematics Calculus

Given:

$$\frac{dy}{dx}+\frac{5}{x(1+{x}^{5})}y=\frac{{(1+{x}^{5})}^{2}}{{x}^{7}}$$dxdy​+x(1+x5)5​y=x7(1+x5)2​

This is linear differential equation.

$$I.F.={e}^{\int \frac{5}{x(1+{x}^{5})}dx}$$I.F.=e∫x(1+x5)5​dx

$$\Rightarrow I.F.={e}^{\int \frac{5{x}^{5}}{{x}^{6}(1+{x}^{5})}dx}$$⇒I.F.=e∫x6(1+x5)5x5​dx

$$\Rightarrow I.F.={e}^{\int \frac{5}{{x}^{6}({x}^{-5}+1)}dx}$$⇒I.F.=e∫x6(x−5+1)5​dx

$$\Rightarrow I.F.={e}^{-\int \frac{-5}{{x}^{6}({x}^{-5}+1)}dx}$$⇒I.F.=e−∫x6(x−5+1)−5​dx

$$\Rightarrow I.F.={e}^{-\int \frac{d({x}^{-5}+1)}{({x}^{-5}+1)}}$$⇒I.F.=e−∫(x−5+1)d(x−5+1)​

$$\Rightarrow I.F.={e}^{-{\mathrm{log}}_{e}({x}^{-5}+1)}$$⇒I.F.=e−loge​(x−5+1)

$$\Rightarrow I.F.={({x}^{-5}+1)}^{-1}=\frac{{x}^{5}}{({x}^{5}+1)}$$⇒I.F.=(x−5+1)−1=(x5+1)x5​

So, solution is given by

$$y\frac{{x}^{5}}{({x}^{5}+1)}=\int \frac{{x}^{5}}{({x}^{5}+1)}\times \frac{{({x}^{5}+1)}^{2}}{{x}^{7}}dx$$y(x5+1)x5​=∫(x5+1)x5​×x7(x5+1)2​dx

$$\Rightarrow \frac{y{x}^{5}}{({x}^{5}+1)}=\int (\frac{{x}^{5}+1}{{x}^{2}})dx$$⇒(x5+1)yx5​=∫(x2x5+1​)dx

$$\Rightarrow \frac{y{x}^{5}}{({x}^{5}+1)}=\int ({x}^{3}+\frac{1}{{x}^{2}})dx$$⇒(x5+1)yx5​=∫(x3+x21​)dx

$$\Rightarrow \frac{y{x}^{5}}{({x}^{5}+1)}=\frac{{x}^{4}}{4}-\frac{1}{x}+C$$⇒(x5+1)yx5​=4x4​−x1​+C

Since, $y(1)=2$y(1)=2, so

$$\frac{2}{2}=\frac{1}{4}-1+C\Rightarrow C=\frac{7}{4}$$22​=41​−1+C⇒C=47​

So,

$$\frac{y{x}^{5}}{({x}^{5}+1)}=\frac{{x}^{4}}{4}-\frac{1}{x}+\frac{7}{4}$$(x5+1)yx5​=4x4​−x1​+47​

Put $x=2$x=2, then we get

$$\frac{y(2)\times 32}{33}=\frac{16}{4}-\frac{1}{2}+\frac{7}{4}$$33y(2)×32​=416​−21​+47​

$$\Rightarrow y(2)=\frac{21\times 33}{128}=\frac{693}{128}$$⇒y(2)=12821×33​=128693​