Let y=y(x) be the solution of the differential equation (1+x^2) (d y)/(d x)+y=e^ ^ -1 x , y(1)=0 . Then y(0) is

Correct answer: B. (1)/(2)(1-e^ π / 2 ) aligned & (d y)/(d x)+(y)/(1+x^2)= e^ ^ -1 x 1+x^2 & I.F. =e^ ∫ (1)/(1+x^2) d x =e^ ^ -1 x & y · e^ ^ -1 x =∫( e^ ^ -1 x 1+x^2 ) e^ ^ -1 x · d x & Let ^ -1 x=z (d x)/(1+x^2)=d z & y · e^z=∫ e^ 2 z d z= e^ 2 z 2 +C & y · e^ ^ -1 x = e^ 2 ^ -1 x 2 +C & y= e^ ^ -1 x 2 + C e^ ^ -1 x aligned aligned & y(1)=0 0= e^ π / 4 2 + C e^ π / 4 C= -e^ π / 2 2 & y= e^ ^ -1 x 2 - e^ π / 2 2 e^ ^ -1 x & y(0)= 1-e^ π / 2 2 aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (1+x^2) (d...

Let

y=y(x)

be the solution of the differential equation \left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}

,

y(1)=0

. Then

y(0)$ is

Options

A $\frac{1}{2}\left(e^{\pi / 2}-1\right)$
B $\frac{1}{2}\left(1-e^{\pi / 2}\right)$
C $\frac{1}{4}\left(1-e^{\pi / 2}\right)$
D $\frac{1}{4}\left(e^{\pi / 2}-1\right)$

Correct answer

B. $\frac{1}{2}\left(1-e^{\pi / 2}\right)$

Step-by-step solution

\begin{aligned} & \frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+x^2} \\ & \text{ I.F. }=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x} \\ & y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^2}\right) e^{\tan ^{-1} x} \cdot d x \\ & \text{ Let } \tan ^{-1} x=z \quad \therefore \frac{d x}{1+x^2}=d z \\ & \therefore y \cdot e^z=\int e^{2 z} d z=\frac{e^{2 z}}{2}+C \\ & y \cdot e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C \\ & \Rightarrow y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}\end{aligned}

\begin{aligned} & \because \mathrm{y}(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2} \\ & \therefore \mathrm{y}=\frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{2}-\frac{\mathrm{e}^{\pi / 2}}{2 \mathrm{e}^{\tan ^{-1} \mathrm{x}}} \\ & \therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}\end{aligned}

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