JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0

. If

y (0) = 0

, then

y (2)

is equal to

Answer: (B)

Solution

(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0 \Rightarrow \frac{d y}{d x} + \frac{2 x ( x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}} y = \frac{2}{( x ^{2} + 4 ) ^{2}} \Rightarrow \frac{d y}{d x} + \frac{2 x}{x ^{2} + 4} y = \frac{2}{( x ^{2} + 4 ) ^{2}}

Integrating factor

= e^{\int \frac{2 x}{x ^{2} + 4} d x} = x^{2} + 4

\frac{d}{d x} [y (x^{2} + 4)] = \frac{2}{x ^{2} + 4} \Rightarrow y (x^{2} + 4) = 2 \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + C = tan^{- 1} \frac{x}{2} + C

y (0) = 0 \Rightarrow C = 0

. At

x = 2

y (8) = tan^{- 1} (1) = \frac{π}{4} \Rightarrow y = \frac{π}{32}

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