JEE Main 2021 — Mathematics Calculus

$${x}^{2}dy+ydx=\frac{dx}{x}$$x2dy+ydx=xdx​

$$\Rightarrow \frac{dy}{dx}+\frac{y}{{x}^{2}}=\frac{1}{{x}^{3}}$$⇒dxdy​+x2y​=x31​

$$I.F={e}^{\int \frac{1}{{x}^{2}}dx}={e}^{-\frac{1}{x}}$$I.F=e∫x21​dx=e−x1​

$$\Rightarrow y\cdot {e}^{-\frac{1}{x}}=\int {e}^{-\frac{1}{x}}\cdot \frac{1}{{x}^{3}}dx+C$$⇒y⋅e−x1​=∫e−x1​⋅x31​dx+C

$$\text{ Let }\frac{-1}{x}=t\Rightarrow \frac{1}{{x}^{2}}dx=dt$$ Let x−1​=t⇒x21​dx=dt

$$\Rightarrow y\cdot {e}^{-\frac{1}{x}}=\int -t{e}^{t}\cdot dt+C$$⇒y⋅e−x1​=∫−tet⋅dt+C

$$=-[t{e}^{t}-{e}^{t}]+C$$=−[tet−et]+C

$$\Rightarrow y\cdot {e}^{-\frac{1}{x}}=\frac{1}{x}{e}^{-\frac{1}{x}}+{e}^{-\frac{1}{x}}+C$$⇒y⋅e−x1​=x1​e−x1​+e−x1​+C

$$\text{ Put }x=1$$ Put x=1

$$\Rightarrow (1)\cdot {e}^{-1}=\frac{{e}^{-1}}{1}+{e}^{-1}+C$$⇒(1)⋅e−1=1e−1​+e−1+C

$$\Rightarrow C=-{e}^{-1}$$⇒C=−e−1

Equation is $y\cdot {e}^{-\frac{1}{x}}=\frac{1}{x}{e}^{-\frac{1}{x}}+{e}^{-\frac{1}{x}}-{e}^{-1}$y⋅e−x1​=x1​e−x1​+e−x1​−e−1

$$\Rightarrow y=\frac{1}{x}+1-\frac{{e}^{\frac{1}{x}}}{e}$$⇒y=x1​+1−eex1​​

$$\text{ At }x=\frac{1}{2}\Rightarrow y(\frac{1}{2})=2+1-\frac{{e}^{2}}{e}\Rightarrow y=3-e$$ At x=21​⇒y(21​)=2+1−ee2​⇒y=3−e