QJEEAdv15Mad005: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Mad005 Solution: Let be a solution of the differential equation If , then which…

Given: $\left(1+e^x\right) y^{\prime}+y e^x=1$

$\left(1+e^x\right) \frac{d y}{d x}+y e^x=1$

$\frac{d y}{d x}+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$ , which is liner differential equation in $y$ .

Comparing above equation with $\frac{d y}{d x}+\mathrm{P} y=Q$ , we get $Q$ , we get

$\begin{aligned}\quad \mathrm{P} & =\frac{e^x}{1+e^x} \text { and } \mathrm{Q}=\frac{1}{1+e^x} \\\text { So, I.F. } & =e^{\int \mathrm{P} \cdot d x} \\\text { I.F. } & =e^{\int \frac{e^x}{1+e^x} d x} \\\text { I.F. } & =e^{\ln \left(1+e^x\right)}\end{aligned}$

$\left\{\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right\}$

$\text { I.F }=1+e^x$

So, solution of given differential equation is given by

$y \text { (I.F.) }=\int \text { Q.(I.F.) } d x$

$\begin{aligned}y \cdot\left(1+e^x\right) & =\int \frac{1}{1+e^x} \cdot\left(1+e^x\right) d x \\y\left(1+e^x\right) & =\int 1 d x \\y\left(1+e^x\right) & =x+\mathrm{C} \\\because \quad y(0) & =2 \\\Rightarrow \quad 2\left(1+e^0\right) & =0+\mathrm{C} \\c & =4\end{aligned}$

So, $y(x)=\frac{x+4}{1+e^x}\quad \text{... (i)}$

Put $x=-4$ in the equation, we get

$y(-4)=0$

Put $x=-2$ in the above equation (i), we get

$y(-2)=\frac{2}{1+e^{-2}} \neq 0$

For critical points, $y^{\prime}=0$

From e.q, (i), $y\left(1+e^x\right)=x+4$

Differentiating the above equation w.r.t. $x$ , we get

$\begin{aligned}y^{\prime}\left(1+e^x\right)+y\left(e^x\right) & =1 \\0\left(1+e^x\right)+y e^x & =1 \quad \left\{\because y^{\prime}=0\right\} \\y e^x-1 & =0\end{aligned}$

$\quad \begin{aligned}\text { Now, } \text { let } g(x) & =y e^x-1 \\g(x) & =\frac{(x+4) e^x}{1+e^x}-1 \\g(x) & =\frac{(x+3) e^x-1}{1+e^x}\end{aligned}$

\text { Now, } \quad g(-1)=\frac{2 e^{-1}-1}{1+e^{-1}}=\frac{2-e}{1+e}<0 \quad \{\because e=2 \cdot 7 \cdot 8\}

\text { And, } \quad g(0)=\frac{3 e^0-1}{1+e^0}=1>0

So, there exists one value of $x$ in $(-1,0)$ for which

$\begin{aligned}g(x) & =0 \\\Rightarrow \quad y^{\prime} & =0\end{aligned}$

There exist a critical point of $y(x)$ in $(-1,0)$

Hint :

(i) Use solution of liner differential equation

$\begin{aligned}\frac{d y}{d x}+\mathrm{P} y & =\mathrm{Q} \text { is given by } \\y \text { (I.F.) } & =\int \mathrm{Q} \cdot \text { (I.F) } d x, \text { where I.F }=e^{\int p \cdot d x}\end{aligned}$

(ii) For critical points, $y^{\prime}=0$