Integrating factor of the differential equation ≤ft(1+x²) d y/d x+x y=1 is

Correct Answer is : 1+x2{1+x²}1+x2

Correct Answer is : 1+x2{1+x²}1+x2

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KCET 2026 Mathematics Solved Paper

Question : 47 of 60

Marks: +1, -0

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

Solution:

Concept:

For a linear differential equation of the form

\frac{dy}{dx} + P y = Q

, the integrating factor is given by

\text{I.F.} = e^{\int P \, dx}

Explanation:

The given equation is

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

Divide the entire equation by

1+x^{2}

to express it in standard linear form.

We get

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

Hence,

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

Now compute

\int P \, dx = \int \frac{x}{1+x^{2}} \, dx

Let

t = 1+x^{2}

, so

dt = 2x \, dx

Thus,

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

Therefore,

\text{I.F.} = e^{\frac{1}{2} \log\left(1+x^{2}\right)}

Using

e^{\log a} = a

, we get

\text{I.F.} = \left(1+x^{2}\right)^{1/2} = \sqrt{1+x^{2}}

This matches option D.

Answer:

\sqrt{1+x^{2}}

(Option D)

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