Visual SolutionStandard

Question

The general solution of

(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}

is:

(A)

y(1+x^2) = \tan^{-1}x + C

(B)

y(1+x^2) = \sin^{-1}x + C

(C)

y = \frac{\tan^{-1}x}{1+x^2} + C

(D)

y(1+x^2)^2 = \tan^{-1}x + C

Solution Path

Standard form with

P = 2x/(1+x^2)

, IF

= 1+x^2

, integrate to get

y(1+x^2) = \tan^{-1}x + C

01Question Setup

1/4

Solve

(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}

\frac{dy}{dx} + \frac{2x}{1+x^2} y = \frac{1}{(1+x^2)^2}

02Standard Form

2/4

Divide by

(1+x^2)

\frac{dy}{dx} + \frac{2x}{1+x^2} y = \frac{1}{(1+x^2)^2}

. This is linear with

P = \frac{2x}{1+x^2}

P = \frac{2x}{1+x^2}

Q = \frac{1}{(1+x^2)^2}

03Integrating FactorKEY INSIGHT

3/4

= e^{\int P\,dx} = e^{\int \frac{2x}{1+x^2}dx} = e^{\log(1+x^2)} = 1+x^2

. Multiply through:

\frac{d}{dx}[y(1+x^2)] = \frac{1}{1+x^2}

= 1+x^2

04Final Answer

4/4

Integrate:

y(1+x^2) = \tan^{-1}x + C

\boxed{y(1+x^2) = \tan^{-1}x + C}

Concepts from this question2 concepts unlocked

★

STANDARD

Multiply by e^(integral P dx) to make the LHS an exact derivative of y times the integrating factor

\text{IF} = e^{\int P(x)\,dx}, \quad \frac{d}{dx}[y \cdot \text{IF}] = Q(x) \cdot \text{IF}

Converts any first-order linear ODE into a directly integrable form, making it the most widely tested technique in JEE

Linear first-order ODEsNewton's law of coolingRL circuit problems

EASY

Rearrange the equation so that all y terms are on one side and all x terms on the other, then integrate both sides

\int \frac{dy}{g(y)} = \int f(x)\,dx + C

The simplest and most direct solving method; recognising when an equation is separable saves significant time

Separable ODEsPopulation modelsRate of change problems

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