JEE Main 2023 — Mathematics Calculus

Given differential equation is

$${x}^{3}dy+(xy-1)dx=0$$x3dy+(xy−1)dx=0

$$\Rightarrow \frac{dy}{dx}=\frac{1-xy}{{x}^{3}}$$⇒dxdy​=x31−xy​

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

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

This is a linear differential equation.

If $\displaystyle \mathrm{IF}={e}^{\int \frac{1}{{x}^{2}}dx}={e}^{-\frac{1}{x}}$IF=e∫x21​dx=e−x1​

Solution of a given linear differential equation is

$$y{e}^{-\frac{1}{x}}=\int {e}^{-\frac{1}{x}}(\frac{1}{{x}^{3}})dx$$ye−x1​=∫e−x1​(x31​)dx

Put $-\frac{1}{x}=t\Rightarrow \frac{dx}{{x}^{2}}=dt$−x1​=t⇒x2dx​=dt

$$\Rightarrow y{e}^{-\frac{1}{x}}=-\int {e}^{-t}tdt$$⇒ye−x1​=−∫e−ttdt

$$\Rightarrow y{e}^{-\frac{1}{x}}=t{e}^{-t}+{e}^{-t}+C$$⇒ye−x1​=te−t+e−t+C

$$\Rightarrow y=t+1+C{e}^{\frac{1}{x}}$$⇒y=t+1+Cex1​

$$\Rightarrow y=\frac{1}{x}+1+C{e}^{\frac{1}{x}}$$⇒y=x1​+1+Cex1​

Put $x=\frac{1}{2}$x=21​, then

$$3-e=2+1+C{e}^{2}$$3−e=2+1+Ce2

$$C=-\frac{1}{e}$$C=−e1​

So,

$$y=\frac{1}{x}+1-\frac{1}{e}({e}^{\frac{1}{x}})$$y=x1​+1−e1​(ex1​)

So,

$$y(1)=1+1-1=1$$y(1)=1+1−1=1