JEE Mathematics 2026 Question with Solution

Given: $(1+x^2)dy+(y-\tan^{-1}x) \, dx=0$ with $y(0)=1$.

Find: $y(1)$.

Write the equation in linear form:

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

so

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

Thus,

$$P(x)=\frac{1}{1+x^2}, \qquad Q(x)=\frac{\tan^{-1}x}{1+x^2}$$

The integrating factor is

$$\text{I.F.}=e^{\int \frac{1}{1+x^2} \, dx}=e^{\tan^{-1}x}$$

Hence,

$$y \, e^{\tan^{-1}x}=\int \frac{\tan^{-1}x}{1+x^2} e^{\tan^{-1}x} \, dx + C$$

Let

$$t=\tan^{-1}x$$

Then

$$dt=\frac{1}{1+x^2} \, dx$$

So the integral becomes

$$\int t e^t \, dt$$

Using integration by parts,

$$\int t e^t \, dt = t e^t - \int e^t \, dt = (t-1)e^t$$

Substituting back,

$$y \, e^{\tan^{-1}x}=(\tan^{-1}x-1)e^{\tan^{-1}x}+C$$

Therefore,

$$y(x)=\tan^{-1}x-1+Ce^{-\tan^{-1}x}$$

Using $y(0)=1$,

$$1=\tan^{-1}(0)-1+Ce^{-\tan^{-1}(0)}$$ $$1=-1+C$$

so

$$C=2$$

Thus,

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

Now at $x=1$,

$$y(1)=\tan^{-1}(1)-1+2e^{-\tan^{-1}(1)}$$ $$y(1)=\frac{\pi}{4}-1+2e^{-\pi/4}$$ $$y(1)=\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1$$

Therefore, the correct option is B.

The crucial step is to identify the equation as a first-order linear differential equation of the form

$$\frac{dy}{dx}+P(x)y=Q(x)$$

after dividing by $1+x^2$. Once this is done, the integrating factor method gives the particular solution directly. The substitution $t=\tan^{-1}x$ converts the integral into the standard form $\int t e^t \, dt$, after which the initial condition determines the constant.