Exam

JEE Main

Year

2024

Date

5 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Differential Equations

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation d y ~d x...

Let $y=y(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0 .$
Then the area enclosed by the curve $f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y-x=4$ is__________

Step-by-step solution

$$\begin{aligned} & I F=e^{\int \frac{2 x}{\left(1+x^2\right)^2} d x}=e^{\frac{-1}{1+x^2}} \\ & y \cdot e^{\frac{-1}{1+x^2}}=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\frac{-1}{1+x^2} d x} \\ & y \cdot e^{\frac{-1}{1+x^2}}=\frac{x^2}{2}+c \\ & (0,0) \Rightarrow C=0 \\ & y(x)=\frac{x^2}{2} e^{\frac{1}{1+x^2}} \\ & f(x)=\frac{x^2}{2}\end{aligned}$$
$A=\int_{-2}^4(x+4)-\frac{x^2}{2} d x=18$