Let y = y(x) be the solution curve of the differential equation (1 + x²)dy + (y − tan⁻¹x)dx = 0, y(0) = 1. Then the value of y(1) is

A first-order linear ODE has the form: $$\frac{dy}{dx} + P(x)\,y = Q(x)$$ How to identify it: Both $y$ and $\frac{dy}{dx}$ appear to the first power only. No terms like $y^2$, $y \cdot \frac{dy}{dx}$, or $\sqrt{y}$.

In this problem, the given equation $(1+x^2)dy + (y – \tan^{-1}x)dx = 0$ is rearranged as: $$\frac{dy}{dx} = \frac{\tan^{-1}x – y}{1+x^2} = \frac{\tan^{-1}x}{1+x^2} – \frac{y}{1+x^2}$$ Moving $y$ term to the left gives the standard form with $P = \frac{1}{1+x^2}$ and $Q = \frac{\tan^{-1}x}{1+x^2}$.

The integrating factor $\mu = e^{\int P\,dx}$ is chosen because: $$\frac{d}{dx}[y \cdot \mu] = \mu\frac{dy}{dx} + y\frac{d\mu}{dx} = \mu\left(\frac{dy}{dx} + Py\right) = \mu \cdot Q$$ This converts the LHS into a perfect derivative — making the equation directly integrable.

Here $\int \frac{dx}{1+x^2} = \tan^{-1}x$, so $\mu = e^{\tan^{-1}x}$.

IF formula: $e^{\int P(x)\,dx}$ $\int \frac{dx}{1+x^2} = \tan^{-1}x + C$

After substituting $t = \tan^{-1}x$, the RHS becomes $\int t\,e^t\,dt$.

Using ILATE rule, take $u = t$ (algebraic) and $dv = e^t\,dt$ (exponential): $$\int t\,e^t\,dt = t\cdot e^t – \int e^t\,dt = te^t – e^t + C = e^t(t-1) + C$$ This is a standard JEE result — memorise it directly: More generally: $\int x\,e^{ax}\,dx = \frac{e^{ax}}{a}\!\left(x – \frac{1}{a}\right) + C$

The substitution works because both $\tan^{-1}x$ and its derivative $\frac{1}{1+x^2}$ appear in the integrand simultaneously: $$\int \frac{\tan^{-1}x}{1+x^2} \cdot e^{\tan^{-1}x}\,dx \xrightarrow{t=\tan^{-1}x} \int t\,e^t\,dt$$ Golden rule: Whenever you see $f(x) \cdot f'(x)$ or $g(f(x)) \cdot f'(x)$ in an integrand, substitution $t = f(x)$ will simplify it perfectly. Here $f(x) = \tan^{-1}x$, $f'(x) = \frac{1}{1+x^2}$.

The general solution $y = \tan^{-1}x – 1 + C\,e^{-\tan^{-1}x}$ has one arbitrary constant $C$. The initial condition pins down $C$ uniquely:

With $y(0) = 0$:   $0 = 0 – 1 + C \Rightarrow C = 1$ → gives $y(1) = \frac{1}{e^{\pi/4}} + \frac{\pi}{4} – 1$
With $y(0) = 1$:   $1 = 0 – 1 + C \Rightarrow C = 2$ → gives $y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} – 1$ ✓

This shows how critically the initial condition affects the final answer. A small change in $y(0)$ changes $C$ and hence $y(1)$ entirely. Always read the initial condition carefully in exam.

❌ Mistake 1: Not converting to standard form before finding IF. Always divide by the coefficient of $dy/dx$ first.

❌ Mistake 2: Misreading the initial condition. Here $y(0) = 1$, not $0$. This changes $C$ from 1 to 2, changing the final answer completely.

❌ Mistake 3: In integration by parts for $\int t\,e^t\,dt$, taking $u = e^t$ and $dv = t\,dt$ — this makes it more complex. Always take the algebraic term as $u$ (ILATE rule).

❌ Mistake 4: Forgetting that $e^{-\pi/4} = \frac{1}{e^{\pi/4}}$, and writing the final answer incorrectly.

❌ Mistake 5: Substituting $x=1$ without knowing $\tan^{-1}(1) = \frac{\pi}{4}$. Standard inverse trig values must be memorised.

$\tan^{-1}(0) = 0$ $\tan^{-1}(1) = \dfrac{\pi}{4}$ $\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$ $\tan^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$ $\sin^{-1}(1) = \dfrac{\pi}{2}$ $\cos^{-1}(1) = 0$

Standard form: $\frac{dy}{dx} + Py = Q$ IF $= e^{\int P\,dx}$ Solution: $y \cdot \text{IF} = \int Q \cdot \text{IF}\,dx + C$ $\int \frac{dx}{1+x^2} = \tan^{-1}x$ $\int t\,e^t\,dt = e^t(t-1)+C$ $e^{-\pi/4} = \frac{1}{e^{\pi/4}}$

First-order linear ODEs appear in every JEE Main session — typically 1 question worth 4 marks. This specific pattern (ODE with $\tan^{-1}x$ leading to IF $= e^{\tan^{-1}x}$) is a favourite. The complete algorithm is fixed:

(1) Standard form → (2) Compute IF → (3) Integrate RHS using substitution + IBP → (4) Apply initial condition → (5) Substitute target $x$ value.

Mastering this 5-step algorithm guarantees full marks. Average time to solve: 3–4 minutes in exam. Practise until each step takes under 45 seconds.