Find the general solution of the differential equation: $\left(x^{2}+1\right)\dfrac{dy}{dx}+2xy=\sqrt{x^{2}+4}$

$\left(x^{2}+1\right)y=\dfrac{x}{2}\sqrt{x^{2}+4}+2\log\left|x+\sqrt{x^{2}+4}\right|+C$

CBSE Class 12 · 2023 · Differential EquationsLong Answer

Find the general solution of the differential equation:

(x^{2} + 1) \frac{d y}{d x} + 2 x y = x^{2} + 4

Answer: $(x^{2} + 1) y = \frac{x}{2} x^{2} + 4 + 2 lo g x + x^{2} + 4 + C$

Divide by

(x^{2} + 1)

\frac{d y}{d x} + \frac{2 x}{x ^{2} + 1} y = \frac{x ^{2} + 4}{x ^{2} + 1}

I.F. = e^{\int \frac{2 x}{x ^{2} + 1} d x} = e^{l o g (x^{2} + 1)} = x^{2} + 1

(x^{2} + 1) y = \int (x^{2} + 1) \cdot \frac{x ^{2} + 4}{x ^{2} + 1} d x = \int x^{2} + 4 d x

Using

\int x^{2} + a^{2} d x = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} lo g x + x^{2} + a^{2}

with

a^{2} = 4

(x^{2} + 1) y = \frac{x}{2} x^{2} + 4 + 2 lo g x + x^{2} + 4 + C

CBSE Class 12 Board · 2023 · Subjective