Let y=y(x) be the solution curve of the differential equation (1+x^2) d y+(y-tan ^-1 x) d x=0, y(0)=1. Then the value of y(1) is :

frac dy dx+frac y x^2+1=fractan ^-1 x x^2+1 I.f. =e^tan ^-1 x y x e^tan ^-1 x=integral e^tan ^-1 x * fractan ^-1 x1+x^2 d x y x e^tan ^-1 x=tan ^-1 x(e^tan ^-1 x)-e^tan ^-1 x+c y(o)=1 => c=2 y(1)=frac2 e^pi / 4+(pi / 4)-1

Let y y x be the solution curve of the differential equ... | null (Mathematics)

JEE Mains 2026 21st January Shift 1

Mathematics

Let

y=y(x)

be the solution curve of the differential equation

\left(1+x^{2}\right) d y+\left(y-\tan ^{-1} x\right) d x=0

y(0)=1

. Then the value of

y(1)

is :

\frac{2}{e^{\frac{\pi}{4}}}+\frac{\pi}{4}-1

\frac{2}{e^{\frac{\pi}{4}}}-\frac{\pi}{4}-1

\frac{4}{e^{\frac{\pi}{4}}}+\frac{\pi}{2}-1

\frac{4}{e^{\frac{\pi}{4}}}-\frac{\pi}{2}-1

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Explanation

\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}^{2}+1}=\frac{\tan ^{-1} \mathrm{x}}{\mathrm{x}^{2}+1}\\

I.f.

=e^{\tan ^{-1} x}\\

y \times e^{\tan ^{-1} x}=\int e^{\tan ^{-1} x} \cdot \frac{\tan ^{-1} x}{1+x^{2}} d x\\

y \times e^{\tan ^{-1} x}=\tan ^{-1} x\left(e^{\tan ^{-1} x}\right)-e^{\tan ^{-1} x}+c\\

y(o)=1 \Rightarrow c=2\\

\mathrm{y}(1)=\frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1

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