JEE Main 2020 — Mathematics Calculus

Given ${x}^{3}dy+xy\cdot dx={x}^{2}dy+2ydx$x3dy+xy⋅dx=x2dy+2ydx

$$\Rightarrow ({x}^{3}-{x}^{2})dy=(2-x)ydx$$⇒(x3−x2)dy=(2−x)ydx

$$\Rightarrow \frac{dy}{y}=\frac{(2-x)}{({x}^{3}-{x}^{2})}dx$$⇒ydy​=(x3−x2)(2−x)​dx

Integrating both sides with respect to $x$x, we get

$$\int \frac{dy}{y}=\int \frac{(2-x)}{{x}^{2}(x-1)}dx+k............(i)$$∫ydy​=∫x2(x−1)(2−x)​dx+k............(i)

Let $\frac{(2-x)}{{x}^{2}(x-1)}=\frac{A}{x}+\frac{B}{{x}^{2}}+\frac{C}{(x-1)}$x2(x−1)(2−x)​=xA​+x2B​+(x−1)C​

$$\Rightarrow (2-x)=Ax(x-1)+B(x-1)+C{x}^{2}$$⇒(2−x)=Ax(x−1)+B(x−1)+Cx2

Putting $x=0\Rightarrow 2=-B\Rightarrow B=-2$x=0⇒2=−B⇒B=−2

Putting $x=1\Rightarrow 2-1=C\Rightarrow C=1$x=1⇒2−1=C⇒C=1

Putting $x=2\Rightarrow 2-2=A(2)(1)+B(1)+C({2}^{2})\Rightarrow 2A+2=0\Rightarrow A=-1$x=2⇒2−2=A(2)(1)+B(1)+C(22)⇒2A+2=0⇒A=−1

From equation $(i)$(i), we get

$$\int \frac{dy}{y}=\int (\frac{-1}{x}+\frac{-2}{{x}^{2}}+\frac{1}{x-1})dx+k$$∫ydy​=∫(x−1​+x2−2​+x−11​)dx+k

$$\Rightarrow \mathrm{ln}y=-\mathrm{ln}x+\frac{2}{x}+\mathrm{ln}|x-1|+k..........(ii)$$⇒lny=−lnx+x2​+ln∣x−1∣+k..........(ii)

Given $y(2)=e$y(2)=e i.e. at $x=2,y=e$x=2,y=e

$$\Rightarrow \mathrm{ln}e=-\mathrm{ln}2+\frac{2}{2}+\mathrm{ln}|2-1|+k$$⇒lne=−ln2+22​+ln∣2−1∣+k

$$\Rightarrow 1=-\mathrm{ln}2+1+0+k$$⇒1=−ln2+1+0+k

$$\Rightarrow k=\mathrm{ln}2$$⇒k=ln2

Putting in equation $(ii)$(ii), we get

$$\mathrm{ln}y=-\mathrm{ln}x+\frac{2}{x}+\mathrm{ln}|x-1|+\mathrm{ln}2$$lny=−lnx+x2​+ln∣x−1∣+ln2

Now putting $x=4$x=4, we get

$$\mathrm{ln}y=-\mathrm{ln}4+\frac{2}{4}+\mathrm{ln}|4-1|+\mathrm{ln}2$$lny=−ln4+42​+ln∣4−1∣+ln2

$$\Rightarrow \mathrm{ln}y=-2\mathrm{ln}2+\frac{1}{2}+\mathrm{ln}3+\mathrm{ln}2$$⇒lny=−2ln2+21​+ln3+ln2

$$\Rightarrow \mathrm{ln}y=-\mathrm{ln}2+\frac{1}{2}+\mathrm{ln}3$$⇒lny=−ln2+21​+ln3

$$\Rightarrow \mathrm{ln}y=\frac{1}{2}+\mathrm{ln}\frac{3}{2}$$⇒lny=21​+ln23​

$$\Rightarrow \mathrm{ln}\frac{2y}{3}=\frac{1}{2}$$⇒ln32y​=21​

$$\Rightarrow \frac{2y}{3}={e}^{\frac{1}{2}}$$⇒32y​=e21​

$$\Rightarrow y=\frac{3}{2}\sqrt{e}$$⇒y=23​e​

$$\Rightarrow y(4)=\frac{3}{2}\sqrt{e}$$⇒y(4)=23​e​