Let the Solution Curve y = y(x) of dy/dx − 3x⁵tan⁻¹(x³)y/(1+x⁶)^(3/2) = 2x·exp((x³−tan⁻¹x³)/√(1+x⁶)) Pass Through the Origin. Then y(1) Is Equal To

JEE Main 2023 — 31 January, Evening Shift. Previous Year Question.

Problem

Let the solution curve $y = y(x)$ of the differential equation

$$\frac{dy}{dx} – \frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\,y = 2x\exp\!\left(\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)$$

pass through the origin. Then $y(1)$ is equal to:

Key insight. The equation looks intimidating only because of the $\tan^{-1}(x^3)$ inside the coefficient of $y$. Substituting $u = x^3$ turns the integrating factor’s exponent into a standard $\int \dfrac{u\tan^{-1}u}{(1+u^2)^{3/2}}\,du$, which integrates by parts cleanly — and once you multiply through, the exponential on the right-hand side is designed to cancel the integrating factor almost entirely, leaving just $2x$.

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Approach

This is a first-order linear differential equation in $y$, so the natural route is the integrating factor method: write it as $\dfrac{dy}{dx} + P(x)y = Q(x)$, find $\mu = e^{\int P\,dx}$, and then $y\mu = \int Q\mu\,dx$. The substitution $u=x^3$ is what makes the integral for $\mu$ tractable, and the same substitution reveals that $Q(x)\cdot\mu(x)$ collapses to something as simple as $2x$ — a strong hint that the problem was built backwards from a clean answer.

Solution

Step 1 — Identify P(x) and set up the integrating-factor integral

Comparing with $\dfrac{dy}{dx}+P(x)y=Q(x)$:

$$P(x) = -\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}, \qquad Q(x) = 2x\exp\!\left(\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)$$

We need $\int P(x)\,dx = -\displaystyle\int \frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\,dx$.

Step 2 — Substitute u = x³ and integrate by parts

Let $u=x^3$, so $du = 3x^2\,dx$, and hence $3x^5\,dx = x^3(3x^2\,dx) = u\,du$. Also $1+x^6 = 1+u^2$. The integral becomes

$$I = \int \frac{u\tan^{-1}u}{(1+u^2)^{3/2}}\,du$$

Take $f = \tan^{-1}u$ and $dg = \dfrac{u}{(1+u^2)^{3/2}}\,du$. Since $\dfrac{d}{du}\left(\dfrac{-1}{\sqrt{1+u^2}}\right) = \dfrac{u}{(1+u^2)^{3/2}}$, we get $g = -\dfrac{1}{\sqrt{1+u^2}}$. Integrating by parts:

$$I = -\frac{\tan^{-1}u}{\sqrt{1+u^2}} + \int \frac{1}{(1+u^2)^{3/2}}\,du = -\frac{\tan^{-1}u}{\sqrt{1+u^2}} + \frac{u}{\sqrt{1+u^2}}$$

using the standard result $\displaystyle\int \frac{du}{(1+u^2)^{3/2}} = \frac{u}{\sqrt{1+u^2}}$. So

$$I = \frac{u-\tan^{-1}u}{\sqrt{1+u^2}} \implies \int P(x)\,dx = -\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}$$

Step 3 — Write the integrating factor and watch the right-hand side collapse

$$\mu(x) = \exp\!\left(-\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)$$

Now multiply $Q(x)$ by $\mu(x)$ — the exponent in $Q$ is exactly the negative of the exponent in $\mu$, so they cancel:

$$Q(x)\cdot\mu(x) = 2x\exp\!\left(\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)\cdot\exp\!\left(-\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right) = 2x$$

This is exactly why the problem was set up this way: no matter how messy $\mu(x)$ looks, the product $Q\mu$ is just $2x$.

Step 4 — Integrate and apply the condition at the origin

$$\frac{d}{dx}(y\mu) = 2x \implies y\mu = x^2 + C$$

Since the curve passes through the origin, $y(0)=0$. At $x=0$, $\mu(0) = \exp(0) = 1$, so $0 = 0 + C$, giving $C=0$. Hence

$$y\cdot\exp\!\left(-\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right) = x^2 \implies y = x^2\exp\!\left(\frac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)$$

Step 5 — Evaluate at x = 1

At $x=1$: $x^3=1$, $\tan^{-1}1 = \dfrac{\pi}{4}$, and $\sqrt{1+1^6}=\sqrt2$. So

$$y(1) = 1\cdot\exp\!\left(\frac{1-\pi/4}{\sqrt2}\right) = \exp\!\left(\frac{4-\pi}{4\sqrt2}\right)$$

Answer

Common mistakes

• Trying to integrate $\int\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\,dx$ directly in terms of $x$. It looks like a dead end until the substitution $u=x^3$ turns it into a clean integration-by-parts problem — spotting that substitution early saves a lot of algebra.
• Not noticing that $Q(x)$ was built to cancel $\mu(x)$. Once $\int P\,dx$ is found, it’s worth immediately comparing its negative to the exponent inside $Q(x)$ — many JEE differential-equation problems are constructed exactly this way, and recognising the pattern turns a hard integral into a two-line simplification.
• Forgetting $\mu(0)=1$. Since the exponent of $\mu$ vanishes at $x=0$ (both $x^3$ and $\tan^{-1}x^3$ are $0$), $\mu(0)=e^0=1$ — easy to overlook when the expression for $\mu$ looks complicated.

Practise next

• Solve $\dfrac{dy}{dx}+\dfrac{2x}{1+x^4}\,y=\dfrac{2x}{1+x^4}$ with $y(0)=0$, and find $y(1)$.