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Differential Equations: Mathematics | JEE Main

Q: Why does simplify so cleanly?

The examiner deliberately engineered to contain the exact reciprocal exponential factor of the integrating factor, so their product is , leaving just the polynomial term .

Let the solution curve

y = y(x)

of the differential equation

\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{\frac{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 :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\exp\left(\frac{4+\pi}{4\sqrt{2}}\right)

\exp\left(\frac{1-\pi}{4\sqrt{2}}\right)

\exp\left(\frac{\pi-4}{4\sqrt{2}}\right)

\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)

Step-by-step solutionView

Correct answer

The value of $y(1)$ is $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ , found by computing the integrating factor using integration by parts and solving the initial value problem.

Option analysis

Why each option works or fails

A ·

\exp\left(\frac{4+\pi}{4\sqrt{2}}\right)

Believing that the integration by parts or arctangent evaluation yields a positive sign for $\tan^{-1}(1)$ , leading to a sum in the numerator. Retain the minus sign from $-\tan^{-1}(x^3)$ in the exponent when evaluating at $x = 1$ , which gives $\frac{1 - \pi/4}{\sqrt{2}} = \frac{4 - \pi}{4\sqrt{2}}$ .

B ·

\exp\left(\frac{1-\pi}{4\sqrt{2}}\right)

Multiplying the whole numerator by $4$ but forgetting to multiply the leading term $1$ by $4$ when combining $1 - \frac{\pi}{4}$ . Put $1 - \frac{\pi}{4}$ over a common denominator: $1 - \frac{\pi}{4} = \frac{4 - \pi}{4}$ , so dividing by $\sqrt{2}$ yields $\frac{4 - \pi}{4\sqrt{2}}$ rather than $\frac{1 - \pi}{4\sqrt{2}}$ .

C ·

\exp\left(\frac{\pi-4}{4\sqrt{2}}\right)

Inverting the sign of the entire exponent during division by the integrating factor or evaluation. Ensure the signs of both terms in the exponent are tracked carefully; $x^3 - \tan^{-1}(x^3)$ evaluated at $x=1$ is $1 - \frac{\pi}{4} > 0$ , so the exponent is $\frac{4-\pi}{4\sqrt{2}}$ .

D ·

\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)

None. This option correctly evaluates the integrating factor and determines $y(1)$ . Correctly compute the integrating factor $I(x) = \exp\left(-\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ , integrate $2x$ , and apply $y(0)=0$ to find $y(1) = \exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ .

Reviewed route

Solution

StepWorking

01given

The differential equation is $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}y = 2x\exp\left\{\frac{x^3 - \tan^{-1}x^3}{\sqrt{1+x^6}}\right\}$ with initial condition $y(0) = 0$ .

02goal

Find the value of $y(1)$ .

03approach

The equation is linear in $y$ : $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = -\frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}$ . We compute the integrating factor $\text{I.F.} = e^{\int P(x)dx}$ by the trigonometric substitution $\tan^{-1}(x^3) = t$ , solve for $y(x)$ using the initial condition, and evaluate at $x = 1$ .

04execute

Compute $\int P(x) dx = -\int \frac{3x^2 \cdot x^3 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} dx$ . Substitute $t = \tan^{-1}(x^3) \implies x^3 = \tan t$ and $dt = \frac{3x^2}{1+x^6} dx$ . The integral becomes $-\int \frac{t \tan t}{\sqrt{1+\tan^2 t}} dt = -\int t \sin t \, dt$ . Using integration by parts, $-\int t \sin t \, dt = -(-t \cos t + \sin t) = t \cos t - \sin t$ . Substituting back $t = \tan^{-1}(x^3)$ , we have $\cos t = \frac{1}{\sqrt{1+x^6}}$ and $\sin t = \frac{x^3}{\sqrt{1+x^6}}$ , so $\int P(x) dx = \frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}$ . Thus, $\text{I.F.} = \exp\left(\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}\right)$ .

05execute

Multiply both sides by $\text{I.F.}$ : $y \cdot \text{I.F.} = \int Q(x) \cdot \text{I.F.} \, dx$ . Notice $Q(x) = 2x \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ , so $Q(x) \cdot \text{I.F.} = 2x \exp(0) = 2x$ . Therefore, $y \exp\left(\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}\right) = \int 2x \, dx = x^2 + C$ .

06execute

Apply the initial condition $y(0) = 0$ : $0 \cdot \exp(0) = 0^2 + C \implies C = 0$ . Hence, $y(x) = x^2 \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ .

07execute

Substitute $x = 1$ : $y(1) = 1^2 \cdot \exp\left(\frac{1 - \tan^{-1}(1)}{\sqrt{1+1^6}}\right) = \exp\left(\frac{1 - \frac{\pi}{4}}{\sqrt{2}}\right) = \exp\left(\frac{4 - \pi}{4\sqrt{2}}\right)$ .

✓verify

Check at $x=0$ : $y(0) = 0 \cdot e^0 = 0$ , satisfying the initial condition. For $x>0$ , $x^3 > \tan^{-1}(x^3)$ , so the exponent $\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} > 0$ . At $x=1$ , exponent is $\frac{4-\pi}{4\sqrt{2}} > 0$ since $\pi < 4$ , which matches option (3).

Hints that build this answer step by step

The differential equation is in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = -\frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}$ . What is the integral $\int P(x)\,dx$ needed for the integrating factor?
$-\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}$
Using the integrating factor $I(x) = \exp\left(-\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ , what is the simplified equation for $y(x) \cdot I(x)$ before integration?
$\frac{d}{dx}[y \cdot I(x)] = 2x$
Integrating both sides and applying the initial condition $y(0) = 0$ , what is the expression for $y(x)$ ?
$y(x) = x^2 \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$
What is the numerical value of $y(1)$ ?
$\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

Q(x) \cdot \text{I.F.}

simplify so cleanly?

The examiner deliberately engineered $Q(x)$ to contain the exact reciprocal exponential factor of the integrating factor, so their product is $1$ , leaving just the polynomial term $2x$ .

Answer

The value of $y(1)$ is $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ , found by computing the integrating factor using integration by parts and solving the initial value problem.

Why each option works or fails

A: $\exp\left(\frac{4+\pi}{4\sqrt{2}}\right)$ - Believing that the integration by parts or arctangent evaluation yields a positive sign for $\tan^{-1}(1)$ , leading to a sum in the numerator. Retain the minus sign from $-\tan^{-1}(x^3)$ in the exponent when evaluating at $x = 1$ , which gives $\frac{1 - \pi/4}{\sqrt{2}} = \frac{4 - \pi}{4\sqrt{2}}$ .
B: $\exp\left(\frac{1-\pi}{4\sqrt{2}}\right)$ - Multiplying the whole numerator by $4$ but forgetting to multiply the leading term $1$ by $4$ when combining $\frac{dy}{dx} + P(x)y = Q(x)$ . Put $1 - \frac{\pi}{4}$ over a common denominator: $1 - \frac{\pi}{4} = \frac{4 - \pi}{4}$ , so dividing by $\sqrt{2}$ yields $\frac{4 - \pi}{4\sqrt{2}}$ rather than $\frac{1 - \pi}{4\sqrt{2}}$ .
C: $\exp\left(\frac{\pi-4}{4\sqrt{2}}\right)$ - Inverting the sign of the entire exponent during division by the integrating factor or evaluation. Ensure the signs of both terms in the exponent are tracked carefully; $x^3 - \tan^{-1}(x^3)$ evaluated at $x=1$ is $1 - \frac{\pi}{4} > 0$ , so the exponent is $\frac{4-\pi}{4\sqrt{2}}$ .
D · correct: $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ - None. This option correctly evaluates the integrating factor and determines $y(1)$ . Correctly compute the integrating factor $I(x) = \exp\left(-\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ , integrate $2x$ , and apply $y(0)=0$ to find $y(1) = \exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ .

Step-by-step solution

given: The differential equation is $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}y = 2x\exp\left\{\frac{x^3 - \tan^{-1}x^3}{\sqrt{1+x^6}}\right\}$ with initial condition $y(0) = 0$ .
goal: Find the value of $y(1)$ .
approach: The equation is linear in $y$ : $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = -\frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}$ . We compute the integrating factor $\text{I.F.} = e^{\int P(x)dx}$ by the trigonometric substitution $\tan^{-1}(x^3) = t$ , solve for $y(x)$ using the initial condition, and evaluate at $x = 1$ .
execute: Compute $\int P(x) dx = -\int \frac{3x^2 \cdot x^3 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} dx$ . Substitute $t = \tan^{-1}(x^3) \implies x^3 = \tan t$ and $dt = \frac{3x^2}{1+x^6} dx$ . The integral becomes $-\int \frac{t \tan t}{\sqrt{1+\tan^2 t}} dt = -\int t \sin t \, dt$ . Using integration by parts, $-\int t \sin t \, dt = -(-t \cos t + \sin t) = t \cos t - \sin t$ . Substituting back $t = \tan^{-1}(x^3)$ , we have $\cos t = \frac{1}{\sqrt{1+x^6}}$ and $\sin t = \frac{x^3}{\sqrt{1+x^6}}$ , so $\int P(x) dx = \frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}$ . Thus, $\text{I.F.} = \exp\left(\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}\right)$ .
execute: Multiply both sides by $\text{I.F.}$ : $y \cdot \text{I.F.} = \int Q(x) \cdot \text{I.F.} \, dx$ . Notice $Q(x) = 2x \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ , so $Q(x) \cdot \text{I.F.} = 2x \exp(0) = 2x$ . Therefore, $y \exp\left(\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}\right) = \int 2x \, dx = x^2 + C$ .
execute: Apply the initial condition $y(0) = 0$ : $0 \cdot \exp(0) = 0^2 + C \implies C = 0$ . Hence, $y(x) = x^2 \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ .
execute: Substitute $x = 1$ : $y(1) = 1^2 \cdot \exp\left(\frac{1 - \tan^{-1}(1)}{\sqrt{1+1^6}}\right) = \exp\left(\frac{1 - \frac{\pi}{4}}{\sqrt{2}}\right) = \exp\left(\frac{4 - \pi}{4\sqrt{2}}\right)$ .
verify: Check at $x=0$ : $y(0) = 0 \cdot e^0 = 0$ , satisfying the initial condition. For $x>0$ , $x^3 > \tan^{-1}(x^3)$ , so the exponent $\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} > 0$ . At $x=1$ , exponent is $\frac{4-\pi}{4\sqrt{2}} > 0$ since $\pi < 4$ , which matches option (3).

Differential Equations: Mathematics | JEE Main

What feels right?

The value of $y(1)$ is $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ , found by computing the integrating factor using integration by parts and solving the initial value problem.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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