JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0\). Then the area enclosed by the curve \(f(\mathrm{x})=\mathrm{y}(\mathrm{x}) \mathrm{e}^{-\frac{1}{\left(1+\mathrm{x}^2\right)}}\) and the line \(\mathrm{y}-\mathrm{x}=4\) is ...........
- A \(62\)
- B \(18\)
- C \(35\)
- D \(16\)
Answer & Solution
Correct Answer
(B) \(18\)
Step-by-step Solution
Detailed explanation
\( \text { IF }=e^{\int \frac{2 x}{\left(1+x^2\right)^2} d x}=e^{\frac{-1}{1+x^2}} \) \( y \cdot e^{\frac{-1}{1+x^2}}=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\frac{-1}{1+x^2} d x} \) \( y \cdot e^{\frac{-1}{1+x^2}}=\frac{x^2}{2}+c \) \( (0,0) \Rightarrow C=0 \)…
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