JEE Main 2024 — Mathematics Calculus

Given: $\frac{dx}{dy}=\frac{1+x-{y}^{2}}{y}$dydx​=y1+x−y2​

$$\Rightarrow \frac{dx}{dy}=\frac{1+x}{y}-y$$⇒dydx​=y1+x​−y

$$\Rightarrow \frac{dx}{dy}-\frac{1+x}{y}=-y$$⇒dydx​−y1+x​=−y

Now let $1+x=t\Rightarrow \frac{dx}{dy}=\frac{dt}{dy}$1+x=t⇒dydx​=dydt​

$$\Rightarrow \frac{dt}{dy}-\frac{t}{y}=-y$$⇒dydt​−yt​=−y

$$\Rightarrow \mathrm{IF}={e}^{\int -\frac{1}{y}dy}$$⇒IF=e∫−y1​dy

$$\Rightarrow \mathrm{IF}={e}^{-\mathrm{log}y}$$⇒IF=e−logy

$$\Rightarrow \mathrm{IF}={e}^{\mathrm{log}\frac{1}{y}}$$⇒IF=elogy1​

$$\Rightarrow \mathrm{IF}=\frac{1}{y}$$⇒IF=y1​

So, the solution of the given differential equation is

$$\frac{t}{y}=\int \frac{1}{y}\times (-y)dy$$yt​=∫y1​×(−y)dy

$$\Rightarrow \frac{(1+x)}{y}=\int \frac{1}{y}\times (-y)dy$$⇒y(1+x)​=∫y1​×(−y)dy

$$\Rightarrow \frac{(1+x)}{y}=-y+c$$⇒y(1+x)​=−y+c

Also, $x(1)=1$x(1)=1

$$\Rightarrow \frac{2}{1}=-1+c$$⇒12​=−1+c

$$\Rightarrow c=3$$⇒c=3

$$\Rightarrow \frac{(1+x)}{y}=-y+3$$⇒y(1+x)​=−y+3

Putting $y=2$y=2

$$\Rightarrow \frac{(1+x)}{2}=-2+3$$⇒2(1+x)​=−2+3

$$\Rightarrow (1+x)=2$$⇒(1+x)=2

$$\Rightarrow x=1$$⇒x=1

$$\Rightarrow 5x=5$$⇒5x=5