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Differential Equations question

2024 · 5 Apr · Shift 2 · Q51
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  5. /2024 · 5 Apr · Shift 2 · Q51

Differential Equations question

2024 · 5 Apr · Shift 2 · Q51

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dy dx+2x(1+x2)2y=xe1(1+x2);y(0)=0.\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. dxdy​+(1+x2)22x​y=xe(1+x2)1​;y(0)=0. Then the area enclosed by the curve f(x)=y(x)e−1(1+x2)f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}f(x)=y(x)e−(1+x2)1​ and the line y−x=4y-x=4y−x=4 is ‾\underline{\hspace{2cm}}​.
Numerical answer
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