JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}

y (1) = 0

. Then

y (0)

Answer: (B)

Solution

= e^{\int \frac{d x}{1 + x ^{2}}} = e^{t a n^{- 1} x}

. Then

(y e^{t a n^{- 1} x})^{'} = \frac{e ^{2 t a n^{- 1} x}}{1 + x ^{2}}

.
With

z = tan^{- 1} x

y e^{t a n^{- 1} x} = \frac{e ^{2 t a n^{- 1} x}}{2} + C

, so

y = \frac{e ^{t a n^{- 1} x}}{2} + C e^{- t a n^{- 1} x}

y (1) = 0

0 = \frac{e ^{π /4}}{2} + C e^{- π /4} \Rightarrow C = - \frac{e ^{π /2}}{2}

y (0) = \frac{1}{2} - \frac{e ^{π /2}}{2} = \frac{1}{2} (1 - e^{π /2}) .

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