JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Solving a homogeneous first-order differential equation using substitution

Q: Why do we divide by x^2 instead of substituting y = vx directly?

Both ways work! Dividing by x^2 directly recognizes d(y/x) = (x dy - y dx)/x^2, which saves 2, 3 lines of algebraic substitution compared to y = vx.

Let the solution curve y=y(x) of the differential equation, [x/(√(x^2-y^2))+e^(y/x)] x dy/dx=x+[x/(√(x^2-y^2))+e^(y/x)] y pass through the points (1,0) and (2 α, α), α>0. Then α is equal to

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A1/2 exp (π/6+√(e)-1)B1/2 exp (π/3+√(e)-1)Cexp (π/6+√(e)+1)D2 exp (π/3+√(e)-1)

Step-by-step solutionView

Correct answer

By letting $y = v x$ , the differential equation separates into $(\frac{1}{\sqrt{1 - v^{2}}} + e^{v}) d v = \frac{d x}{x}$ , yielding $α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$ after using the boundary conditions.

Option analysis

Why each option works or fails

A · 1/2 exp (π/6+√(e)-1)

This is the correct value of $α$ . Correctly integrated $\frac{1}{\sqrt{1 - v^{2}}} + e^{v}$ to $\arcsin (v) + e^{v}$ , applied $(1, 0)$ to get $C = 1$ , evaluated at $(2 α, α)$ where $v = 1 / 2$ , and isolated $α$ .

B · 1/2 exp (π/3+√(e)-1)

The student evaluated $\arcsin (1 / 2)$ as $\frac{π}{3}$ instead of $\frac{π}{6}$ . Recall that $\sin (\frac{π}{6}) = \frac{1}{2}$ , so $\arcsin (\frac{1}{2}) = \frac{π}{6}$ , not $\frac{π}{3}$ .

C · exp (π/6+√(e)+1)

The student solved for $x$ rather than $α$ , missing the factor of $1 / 2$ since $x = 2 α$ , and made a sign error when handling the constant of integration. Ensure you solve for $α$ where $x = 2 α$ , which gives $α = \frac{x}{2}$ , and carefully transfer the constant $C = 1$ across the equality.

D · 2 exp (π/3+√(e)-1)

The student inverted the factor of 2 (multiplying instead of dividing by 2 when $x = 2 α$ ) and also evaluated $\arcsin (1 / 2)$ as $\frac{π}{3}$ . Since $x = 2 α$ , we have $α = \frac{x}{2}$ . Also check trigonometric values: $\arcsin (1 / 2) = \frac{π}{6}$ .

Reviewed route

Solution

StepWorking

01Given

Given differential equation: $[\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}] y$ , passing through $(1, 0)$ and $(2 α, α)$ where $α > 0$ .

02Goal

Find the value of the parameter $α$ .

03Approach

Group the terms with common factor $[\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}]$ together to isolate $(x d y - y d x)$ . Then divide by $x^{2}$ to recognize the exact differential $d (y / x)$ and integrate directly.

04Execute

Rearranging the equation gives: $(\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}) (x d y - y d x) = x d x$ Dividing both sides by $x^{2}$ : $(\frac{1}{\sqrt{1 - (y / x)^{2}}} + e^{y / x}) \frac{x d y - y d x}{x^{2}} = \frac{d x}{x}$ Recognizing $d (\frac{y}{x}) = \frac{x d y - y d x}{x^{2}}$ , the equation becomes: $(\frac{1}{\sqrt{1 - (y / x)^{2}}} + e^{y / x}) d (\frac{y}{x}) = \frac{d x}{x}$

05Execute

Integrate both sides: $\int \frac{1}{\sqrt{1 - (y / x)^{2}}} d (\frac{y}{x}) + \int e^{y / x} d (\frac{y}{x}) = \int \frac{d x}{x}$ $\sin^{- 1} (\frac{y}{x}) + e^{y / x} = \ln ∣ x ∣ + C$

06Execute

Use the condition $(x, y) = (1, 0)$ : $\sin^{- 1} (0) + e^{0} = \ln (1) + C ⟹ 0 + 1 = 0 + C ⟹ C = 1$ Thus, the solution curve is: $\sin^{- 1} (\frac{y}{x}) + e^{y / x} = \ln x + 1 (x > 0)$

07Execute

Substitute $(x, y) = (2 α, α)$ into the curve equation: Here $\frac{y}{x} = \frac{α}{2 α} = \frac{1}{2}$ . $\sin^{- 1} (\frac{1}{2}) + e^{1 / 2} = \ln (2 α) + 1$ $\frac{π}{6} + \sqrt{e} = \ln (2 α) + 1 ⟹ \ln (2 α) = \frac{π}{6} + \sqrt{e} - 1$ $2 α = \exp (\frac{π}{6} + \sqrt{e} - 1) ⟹ α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$

✓Verify

Check at $(2 α, α)$ : $\ln (2 α) = \ln (2) + \ln (α) = \frac{π}{6} + \sqrt{e} - 1$ . Then $\sin^{- 1} (1 / 2) + e^{1 / 2} = π / 6 + \sqrt{e}$ , which perfectly balances $\ln (2 α) + 1$ .

Hints that build this answer step by step

Rearranging the equation yields $[\frac{1}{\sqrt{1 - (y / x)^{2}}} + e^{y / x}] (x d y - y d x) = x d x$ . What standard substitution reduces this equation to separable form?
Substitute $y = v x$ , which gives $d y = v d x + x d v$ and $x d y - y d x = x^{2} d v$ .
After substituting $y = v x$ and dividing by $x^{2}$ , what is the resulting separated differential equation?
$(\frac{1}{\sqrt{1 - v^{2}}} + e^{v}) d v = \frac{d x}{x}$
Integrating both sides gives $\arcsin (v) + e^{v} = \ln ∣ x ∣ + C$ . Using the initial condition that the curve passes through $(1, 0)$ , what is the value of $C$ ?
$C = 1$
Now substitute the point $(x, y) = (2 α, α)$ into the solution $\arcsin (v) + e^{v} = \ln x + 1$ with $v = \frac{y}{x} = \frac{1}{2}$ . What is $α$ ?
$α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why do we divide by

x^{2}

instead of substituting

y = v x

directly?

Both ways work! Dividing by $x^{2}$ directly recognizes $d (y / x) = (x d y - y d x) / x^{2}$ , which saves 2, 3 lines of algebraic substitution compared to $y = v x$ .

Answer

By letting $y = v x$ , the differential equation separates into $(\frac{1}{\sqrt{1 - v^{2}}} + e^{v}) d v = \frac{d x}{x}$ , yielding $α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$ after using the boundary conditions.

Why each option works or fails

A · correct: 1/2 exp (π/6+√(e)-1) - This is the correct value of $α$ . Correctly integrated $\frac{1}{\sqrt{1 - v^{2}}} + e^{v}$ to $\arcsin (v) + e^{v}$ , applied $(1, 0)$ to get $C = 1$ , evaluated at $(2 α, α)$ where $v = 1 / 2$ , and isolated $α$ .
B: 1/2 exp (π/3+√(e)-1) - The student evaluated $\arcsin (1 / 2)$ as $\frac{π}{3}$ instead of $\frac{π}{6}$ . Recall that $\sin (\frac{π}{6}) = \frac{1}{2}$ , so $\arcsin (\frac{1}{2}) = \frac{π}{6}$ , not $\frac{π}{3}$ .
C: exp (π/6+√(e)+1) - The student solved for $x$ rather than $α$ , missing the factor of $1 / 2$ since $x = 2 α$ , and made a sign error when handling the constant of integration. Ensure you solve for $α$ where $x = 2 α$ , which gives $α = \frac{x}{2}$ , and carefully transfer the constant $C = 1$ across the equality.
D: 2 exp (π/3+√(e)-1) - The student inverted the factor of 2 (multiplying instead of dividing by 2 when $x = 2 α$ ) and also evaluated $\arcsin (1 / 2)$ as $\frac{π}{3}$ . Since $x = 2 α$ , we have $α = \frac{x}{2}$ . Also check trigonometric values: $\arcsin (1 / 2) = \frac{π}{6}$ .

Step-by-step solution

Given: Given differential equation: $\frac{π}{6}$ , passing through $(1, 0)$ and $(2 α, α)$ where $α > 0$ .
Goal: Find the value of the parameter $α$ .
Approach: Group the terms with common factor $[\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}]$ together to isolate $x = 2 α$ . Then divide by $x^{2}$ to recognize the exact differential $d (y / x)$ and integrate directly.
Execute: Rearranging the equation gives: $(\frac{x}{\sqrt{x^{2} - y^{2}}} + e^{y / x}) (x d y - y d x) = x d x$ Dividing both sides by $x^{2}$ : $(\frac{1}{\sqrt{1 - (y / x)^{2}}} + e^{y / x}) \frac{x d y - y d x}{x^{2}} = \frac{d x}{x}$ Recognizing $d (\frac{y}{x}) = \frac{x d y - y d x}{x^{2}}$ , the equation becomes: $(\frac{1}{\sqrt{1 - (y / x)^{2}}} + e^{y / x}) d (\frac{y}{x}) = \frac{d x}{x}$
Execute: Integrate both sides: $\int \frac{1}{\sqrt{1 - (y / x)^{2}}} d (\frac{y}{x}) + \int e^{y / x} d (\frac{y}{x}) = \int \frac{d x}{x}$ $\sin^{- 1} (\frac{y}{x}) + e^{y / x} = \ln ∣ x ∣ + C$
Execute: Use the condition $(x, y) = (1, 0)$ : $\sin^{- 1} (0) + e^{0} = \ln (1) + C ⟹ 0 + 1 = 0 + C ⟹ C = 1$ Thus, the solution curve is: $\sin^{- 1} (\frac{y}{x}) + e^{y / x} = \ln x + 1 (x > 0)$
Execute: Substitute $(x, y) = (2 α, α)$ into the curve equation: Here $\frac{y}{x} = \frac{α}{2 α} = \frac{1}{2}$ . $\sin^{- 1} (\frac{1}{2}) + e^{1 / 2} = \ln (2 α) + 1$ $\frac{π}{6} + \sqrt{e} = \ln (2 α) + 1 ⟹ \ln (2 α) = \frac{π}{6} + \sqrt{e} - 1$ $2 α = \exp (\frac{π}{6} + \sqrt{e} - 1) ⟹ α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$
Verify: Check at $(2 α, α)$ : $\ln (2 α) = \ln (2) + \ln (α) = \frac{π}{6} + \sqrt{e} - 1$ . Then $\sin^{- 1} (1 / 2) + e^{1 / 2} = π / 6 + \sqrt{e}$ , which perfectly balances $\ln (2 α) + 1$ .

Solving a homogeneous first-order differential equation using substitution

What feels right?

By letting $y = v x$ , the differential equation separates into $(\frac{1}{\sqrt{1 - v^{2}}} + e^{v}) d v = \frac{d x}{x}$ , yielding $α = \frac{1}{2} \exp (\frac{π}{6} + \sqrt{e} - 1)$ after using the boundary conditions.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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