JEE Main 2024 — Mathematics Calculus

Given: $\frac{dy}{dx}=2x{(x+y)}^{3}-x(x+y)-1$dxdy​=2x(x+y)3−x(x+y)−1

Let, $x+y=t$x+y=t

$$\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$$⇒1+dxdy​=dxdt​

$$\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1$$⇒dxdy​=dxdt​−1

$$\Rightarrow \frac{dt}{dx}-1=2x{t}^{3}-xt-1$$⇒dxdt​−1=2xt3−xt−1

$$\Rightarrow \frac{dt}{dx}=2x{t}^{3}-xt$$⇒dxdt​=2xt3−xt

$$\Rightarrow \frac{dt}{dx}+xt=2x{t}^{3}$$⇒dxdt​+xt=2xt3

$$\Rightarrow \frac{1}{{t}^{3}}\frac{dt}{dx}+\frac{x}{{t}^{2}}=2x...(i)$$⇒t31​dxdt​+t2x​=2x...(i)

Putting, $\frac{1}{{t}^{2}}=u$t21​=u

$$\Rightarrow \frac{-2}{{t}^{3}}\frac{dt}{dx}=\frac{du}{dx}$$⇒t3−2​dxdt​=dxdu​

$$\Rightarrow \frac{1}{{t}^{3}}\frac{dt}{dx}=\frac{-1}{2}\frac{du}{dx}$$⇒t31​dxdt​=2−1​dxdu​

Putting this value in equation $(i)$(i)

$$\Rightarrow \frac{-1}{2}\frac{du}{dx}+xu=2x$$⇒2−1​dxdu​+xu=2x

$$\Rightarrow \frac{du}{dx}-2xu=-4x$$⇒dxdu​−2xu=−4x

Integrating factor, $\displaystyle IF={e}^{\int -2xdx}$IF=e∫−2xdx

$$\Rightarrow IF={e}^{-{x}^{2}}$$⇒IF=e−x2

So, solution of the given differential equation is,

$$u\times {e}^{-{x}^{2}}=\int {e}^{-{x}^{2}}(-4x)dx$$u×e−x2=∫e−x2(−4x)dx

$$\Rightarrow \frac{{e}^{-{x}^{2}}}{{t}^{2}}=\int {e}^{-{x}^{2}}(-4x)dx$$⇒t2e−x2​=∫e−x2(−4x)dx

Putting, $-{x}^{2}=z$−x2=z

$$\Rightarrow -2xdx=dz$$⇒−2xdx=dz

$$\Rightarrow \frac{{e}^{-{x}^{2}}}{{(x+y)}^{2}}=\int 2{e}^{z}dz$$⇒(x+y)2e−x2​=∫2ezdz

$$\Rightarrow \frac{{e}^{-{x}^{2}}}{{(x+y)}^{2}}=2{e}^{z}+C$$⇒(x+y)2e−x2​=2ez+C

$$\Rightarrow \frac{{e}^{-{x}^{2}}}{{(x+y)}^{2}}=2{e}^{-{x}^{2}}+C$$⇒(x+y)2e−x2​=2e−x2+C

$$\Rightarrow \frac{1}{{(x+y)}^{2}}=2+C{e}^{{x}^{2}}$$⇒(x+y)21​=2+Cex2

It is given that, at $x=0$x=0, $y=1$y=1.

$$\Rightarrow 1=2+C$$⇒1=2+C

$$\Rightarrow C=-1$$⇒C=−1

$$\Rightarrow {(x+y)}^{2}=\frac{1}{2-{e}^{{x}^{2}}}$$⇒(x+y)2=2−ex21​

Putting, $x=\frac{1}{\sqrt{2}}$x=2​1​

$$\Rightarrow {(y+\frac{1}{\sqrt{2}})}^{2}=\frac{1}{2-{e}^{\frac{1}{2}}}$$⇒(y+2​1​)2=2−e21​1​

$$\Rightarrow {(y(\frac{1}{\sqrt{2}})+\frac{1}{\sqrt{2}})}^{2}=(\frac{1}{2-\sqrt{e}})$$⇒(y(2​1​)+2​1​)2=(2−e​1​)