JEE Main 2023 — Mathematics Calculus

Given,

$$2(y+2){\mathrm{log}}_{e}(y+2)dx+(x+4-2{\mathrm{log}}_{e}(y+2))dy=0$$2(y+2)loge​(y+2)dx+(x+4−2loge​(y+2))dy=0

Let $x+4=u,y+2=v$x+4=u,y+2=v

$$dx=du,dy=dv$$dx=du,dy=dv

So, the equation becomes,

$$(2v\mathrm{ln}v)\mathrm{du}=-(u-2\mathrm{ln}v)dv$$(2vlnv)du=−(u−2lnv)dv

$$\Rightarrow 2v\mathrm{ln}v\frac{du}{dv}+u=2\mathrm{ln}v$$⇒2vlnvdvdu​+u=2lnv

$$\Rightarrow \frac{du}{dv}+\frac{1}{2v\mathrm{ln}v}.u=\frac{1}{v}$$⇒dvdu​+2vlnv1​.u=v1​

Which is a linear differential equation,

So, $\displaystyle IF={e}^{\frac{1}{2}\int \frac{1}{v\mathrm{ln}v}}={e}^{\frac{1}{2}\mathrm{ln}(\mathrm{ln}v)}={(\mathrm{ln}v)}^{\frac{1}{2}}$IF=e21​∫vlnv1​=e21​ln(lnv)=(lnv)21​

Now solution of differential equation is given by,

$$u\cdot {(\mathrm{ln}v)}^{\frac{1}{2}}=\int \frac{1}{v}\cdot {(\mathrm{ln}v)}^{\frac{1}{2}}dv$$u⋅(lnv)21​=∫v1​⋅(lnv)21​dv

$$\Rightarrow u\cdot {(\mathrm{ln}v)}^{\frac{1}{2}}=\frac{2}{3}{(\mathrm{ln}v)}^{\frac{3}{2}}+c.......(i)$$⇒u⋅(lnv)21​=32​(lnv)23​+c.......(i)

Now using given value, $y={e}^{4}-2\Rightarrow x=1$y=e4−2⇒x=1

$$\therefore v={e}^{4}\Rightarrow u=5$$∴v=e4⇒u=5

$$5\cdot ({4}^{\frac{1}{2}})=\frac{2}{3}\cdot {(4)}^{\frac{3}{2}}+c$$5⋅(421​)=32​⋅(4)23​+c

$$\Rightarrow 10=\frac{16}{3}+c$$⇒10=316​+c

$$\Rightarrow c=\frac{14}{3}$$⇒c=314​

Now finding, $y={e}^{9}-2\Rightarrow v=y+2={e}^{9}$y=e9−2⇒v=y+2=e9

Now putting the value in equation $(i)$(i) we get,

$$\Rightarrow u\cdot 3=\frac{2}{3}\times 27+\frac{14}{3}=18+\frac{14}{3}$$⇒u⋅3=32​×27+314​=18+314​

$$\Rightarrow x+4=u=6+\frac{14}{9}$$⇒x+4=u=6+914​

$$\Rightarrow x=2+\frac{14}{9}=\frac{32}{9}$$⇒x=2+914​=932​