If y = y(x) is the solution of the differential equation xdy \over dx + 2y = x\,e^x , y(1) = 0 then the local maximum value of the function z(x) = x^2y(x) - e^x,\,x \in R is :

Differential Equations PYQ — JEE Main 2022 · 26 Jun · Shift 2 · Q32 with Solution

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Correct answer: D

Given $x\frac{dy}{dx}+2y=xe^x, \qquad y(1)=0.$

For $x\neq 0$ , divide by $x$ : $\frac{dy}{dx}+\frac{2}{x}y=e^x.$

This is a linear differential equation. Its integrating factor is $I.F.=e^{\int \frac{2}{x}\,dx}=e^{2\ln|x|}=x^2.$

Multiplying the equation by $x^2$ , $x^2\frac{dy}{dx}+2xy=x^2e^x.$

The left side is $\frac{d}{dx}(x^2y)=x^2e^x.$

Integrating, $x^2y=\int x^2e^x\,dx + C.$

Now, $\int x^2e^x\,dx=e^x(x^2-2x+2).$

So, $x^2y=e^x(x^2-2x+2)+C.$

Use $y(1)=0$ : $1^2\cdot 0=e(1-2+2)+C=e+C.$ Thus, $C=-e.$

Hence, $x^2y=e^x(x^2-2x+2)-e.$

Given $z(x)=x^2y(x)-e^x.$

Substitute $x^2y$ : $z(x)=\big[e^x(x^2-2x+2)-e\big]-e^x.$

Therefore, $z(x)=e^x(x^2-2x+1)-e=e^x(x-1)^2-e.$

Differentiate: $z'(x)=\frac{d}{dx}\big(e^x(x-1)^2-e\big).$

Using product rule,

=e^x(x-1)\big[(x-1)+2\big].$$ So, $$z'(x)=e^x(x-1)(x+1).$$ Since $e^x>0$, critical points are $$x=1,\quad x=-1.$$ --- 4. **Classify the critical points** Check sign of $z'(x)=e^x(x-1)(x+1)$: - For $x<-1$: $(x-1)(x+1)>0 \Rightarrow z'(x)>0$ - For $-1<x<1$: $(x-1)(x+1)<0 \Rightarrow z'(x)<0$ - For $x>1$: $(x-1)(x+1)>0 \Rightarrow z'(x)>0$ Thus: - at $x=-1$, $z'$ changes from $+$ to $-$, so there is a **local maximum**; - at $x=1$, $z'$ changes from $-$ to $+$, so there is a local minimum. --- 5. **Compute the local maximum value** At $x=-1$, $$z(-1)=e^{-1}(-2)^2-e=\frac{4}{e}-e.$$ So the local maximum value is $$\boxed{\frac{4}{e}-e}.$$ This matches **Option D**.