JEE Main 2022 — Mathematics Calculus

Given $x(1-{x}^{2})\frac{dy}{dx}+(3{x}^{2}y-y-4{x}^{3})=0$x(1−x2)dxdy​+(3x2y−y−4x3)=0

$$(x-{x}^{3})\frac{dy}{dx}+(3{x}^{2}-1)y=4{x}^{3}$$(x−x3)dxdy​+(3x2−1)y=4x3

$$\frac{dy}{dx}+\frac{(3{x}^{2}-1)}{(x-{x}^{3})}y=\frac{4{x}^{3}}{(x-{x}^{3})}$$dxdy​+(x−x3)(3x2−1)​y=(x−x3)4x3​

This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$dxdy​+Py=Q

Here $\displaystyle IF={e}^{\int Pdx}={e}^{\int \frac{3{x}^{2}-1}{x-{x}^{3}}dx}$IF=e∫Pdx=e∫x−x33x2−1​dx

Let $\displaystyle x-{x}^{3}=t\Rightarrow IF={e}^{\int -\frac{\mathrm{dt}}{t}}$x−x3=t⇒IF=e∫−tdt​

$$={e}^{-\mathrm{ln}t}=\frac{1}{t}$$=e−lnt=t1​

$$\therefore IF=\frac{1}{x-{x}^{3}}$$∴IF=x−x31​

So the general solution of the differential equation will be

$$y\times IF=\int Q\times IFdx$$y×IF=∫Q×IFdx

$$y(\frac{1}{x-{x}^{3}})=\int \frac{4{x}^{3}}{x-{x}^{3}}\times \frac{1}{(x-{x}^{3})}dx$$y(x−x31​)=∫x−x34x3​×(x−x3)1​dx

$$=\int \frac{4{x}^{3}}{{(x-{x}^{3})}^{2}}dx$$=∫(x−x3)24x3​dx

$$=\int \frac{4x}{{(1-{x}^{2})}^{2}}dx$$=∫(1−x2)24x​dx

Now let $\displaystyle I=\int \frac{4x}{{(1-{x}^{2})}^{2}}dx$I=∫(1−x2)24x​dx

Substituting $1-{x}^{2}=z$1−x2=z

$$I=2\int \frac{-dz}{{z}^{2}}$$I=2∫z2−dz​

$$=-2(-\frac{1}{z})+c=\frac{2}{1-{x}^{2}}+c$$=−2(−z1​)+c=1−x22​+c

Hence the solution of the differential equation becomes

$$\frac{y}{x-{x}^{3}}=\frac{2}{1-{x}^{2}}+c$$x−x3y​=1−x22​+c

At $x=2,y=-2$x=2,y=−2

$$\frac{-2}{2-8}=\frac{2}{1-4}+c$$2−8−2​=1−42​+c

$$\frac{1}{3}=\frac{-2}{3}+c$$31​=3−2​+c

$$\therefore c=1$$∴c=1

Hence the particular solution will be $\frac{y}{x-{x}^{3}}=\frac{2}{1-{x}^{2}}+1$x−x3y​=1−x22​+1

Put $x=3$x=3

$$\frac{y}{3-27}=\frac{2}{1-9}+1$$3−27y​=1−92​+1

$$\frac{y}{-24}=-\frac{1}{4}+1$$−24y​=−41​+1

$$-\frac{y}{24}=\frac{3}{4}$$−24y​=43​

$$y=\frac{3}{4}(-24)=-18$$y=43​(−24)=−18