JEE MainMathematics

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Solving a First-Order Linear Differential Equation Using an Integrating Factor

Q: Why do we make the substitution t = -1/x instead of direct integration by parts?

Direct integration by parts also works by choosing u = 1/x and dv = 1/(x^2) e^(-1/x) dx, but setting t = -1/x simplifies the integrand to standard polynomial times exponential integral of -t e^t dt, minimizing algebra slips.

Let y=y(x) be the solution of the differential equation x^3 d y+(x y-1) d x=0, x>0, y(1/2)=3-e. Then y(1) is equal to

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A1BeC3D2-e

Step-by-step solutionView

Correct answer

Rewriting the differential equation in standard linear form yields the integrating factor $e^{- 1 / x}$ , giving $y(x) = rac{1}{x} + 1 - e^{2 - 1/x}$ , which evaluates to $1$ at $x = 1$ .

Option analysis

Why each option works or fails

A · 1

This is the correct value obtained by solving the linear differential equation with the given initial condition. Correctly identify the integrating factor $I (x) = e^{- 1 / x}$ , solve for $y (x)$ , use $y (1 / 2) = 3 - e$ to find the constant $C = - e^{2}$ , and evaluate at $x = 1$ .

B · e

A student might omit the constant term or make a sign error when solving for the integration constant $C$ , leading to an extra factor of $e$ remaining. Carefully substitute $x = 1 / 2$ into $y e^{- 1 / x} = (1 / x + 1) e^{- 1 / x} + C$ to ensure $C$ correctly balances with the initial condition $3 - e$ .

C · 3

A student might calculate $C = 0$ incorrectly by misinterpreting $y (1 / 2) = 3 - e$ or dropping the exponential terms entirely. Keep all exponential factors intact when solving $y (1 / 2) e^{- 2} = (2 + 1) e^{- 2} + C$ .

D · 2-e

A student might forget to evaluate the $1 / x$ term at $x = 1$ or subtract the terms incorrectly after evaluating the constant. Ensure that every term of the particular solution $y(x) = rac{1}{x} + 1 - e^{2 - 1/x}$ is computed at $x = 1$ , giving $1 + 1 - e = 2 - e$ only if $C = - e$ rather than $- e^{2}$ .

Reviewed route

Solution

StepWorking

01Given

The differential equation is $x^{3} d y + (x y - 1) d x = 0$ for $x > 0$ , with initial condition $y (\frac{1}{2}) = 3 - e$ .

02Goal

Find the value of $y (1)$ .

03Approach

Rearrange the equation into the standard linear first-order differential equation form $\frac{d y}{d x} + P (x) y = Q (x)$ , determine the integrating factor $I.F. = e^{\int P (x) d x}$ , solve via substitution for the RHS integral, determine the constant $C$ using the initial value, and finally evaluate at $x = 1$ .

04Execute

Rewrite the equation: $x^{3} \frac{d y}{d x} + x y = 1 ⟹ \frac{d y}{d x} + \frac{1}{x^{2}} y = \frac{1}{x^{3}}$ . Here $P (x) = \frac{1}{x^{2}}$ and $Q (x) = \frac{1}{x^{3}}$ . The integrating factor is $I.F. = e^{\int \frac{1}{x^{2}} d x} = e^{- 1 / x}$ . Multiplying gives: $y e^{- 1 / x} = \int \frac{1}{x^{3}} e^{- 1 / x} d x$ . Substitute $t = - \frac{1}{x}$ , so $d t = \frac{1}{x^{2}} d x$ and $\frac{1}{x} = - t$ : $\int \frac{1}{x^{3}} e^{- 1 / x} d x = \int (- t) e^{t} d t = - (t - 1) e^{t} + C = (1 - t) e^{t} + C = (1 + \frac{1}{x}) e^{- 1 / x} + C$ . Hence, $y (x) = 1 + \frac{1}{x} + C e^{1 / x}$ .

05Execute

Use the initial condition $y (\frac{1}{2}) = 3 - e$ : $3 - e = 1 + 2 + C e^{2} = 3 + C e^{2} ⟹ C e^{2} = - e ⟹ C = - e^{- 1} = - \frac{1}{e}$ . Now evaluate at $x = 1$ : $y (1) = 1 + 1 + (- \frac{1}{e}) e^{1} = 2 - 1 = 1$ .

✓Verify

Check at $x = 1 / 2$ : $y (1 / 2) = 1 + 2 - e^{- 1} \cdot e^{2} = 3 - e$ , which matches the given condition. Also, check $y^{'} (1)$ : with $y (1) = 1$ , $y^{'} = 1 - 1 = 0$ ; from general formula $y^{'} = - 1 / x^{2} - e^{1 / x - 1} / x^{2}$ , at $x = 1$ , $y^{'} (1) = - 1 - (- 1) = 0$ . Solution is completely consistent.

Hints that build this answer step by step

How is the given differential equation $x^{3} d y + (x y - 1) d x = 0$ expressed in standard first-order linear form $\frac{d y}{d x} + P (x) y = Q (x)$ ?
$\frac{d y}{d x} + \frac{1}{x^{2}} y = \frac{1}{x^{3}}$
What is the integrating factor $I (x) = e^{\int P (x) d x}$ for this differential equation?
$e^{- 1 / x}$
What is the general solution $y \cdot I (x) = \int Q (x) I (x) d x + C$ ?
$y e^{- 1 / x} = (\frac{1}{x} + 1) e^{- 1 / x} + C$
Using $y (\frac{1}{2}) = 3 - e$ , what is the value of the constant $C$ , and what is $y (1)$ ?
$C = - e^{- 1}$ , leading to $y (1) = 1$

✓ 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: 9 September 2026

Quick checks

Students also ask

Why do we make the substitution

t = - 1 / x

instead of direct integration by parts?

Direct integration by parts also works by choosing $u = 1 / x$ and $d v = \frac{1}{x^{2}} e^{- 1 / x} d x$ , but setting $t = - 1 / x$ simplifies the integrand to standard polynomial times exponential $\int - t e^{t} d t$ , minimizing algebra slips.

Answer

Rewriting the differential equation in standard linear form yields the integrating factor $e^{- 1 / x}$ , giving $y(x) = rac{1}{x} + 1 - e^{2 - 1/x}$ , which evaluates to $1$ at $x = 1$ .

Why each option works or fails

A · correct: 1 - This is the correct value obtained by solving the linear differential equation with the given initial condition. Correctly identify the integrating factor $I (x) = e^{- 1 / x}$ , solve for $y (x)$ , use $y (1 / 2) = 3 - e$ to find the constant $C = - e^{2}$ , and evaluate at $x = 1$ .
B: e - A student might omit the constant term or make a sign error when solving for the integration constant $C$ , leading to an extra factor of $e$ remaining. Carefully substitute $x = 1 / 2$ into $y e^{- 1 / x} = (1 / x + 1) e^{- 1 / x} + C$ to ensure $C$ correctly balances with the initial condition $3 - e$ .
C: 3 - A student might calculate $C = 0$ incorrectly by misinterpreting $y (1 / 2) = 3 - e$ or dropping the exponential terms entirely. Keep all exponential factors intact when solving $y (1 / 2) e^{- 2} = (2 + 1) e^{- 2} + C$ .
D: 2-e - A student might forget to evaluate the $1 / x$ term at $x = 1$ or subtract the terms incorrectly after evaluating the constant. Ensure that every term of the particular solution $y(x) = rac{1}{x} + 1 - e^{2 - 1/x}$ is computed at $x = 1$ , giving $1 + 1 - e = 2 - e$ only if $C = - e$ rather than $- e^{2}$ .

Step-by-step solution

Given: The differential equation is $x^{3} d y + (x y - 1) d x = 0$ for $x > 0$ , with initial condition $y (\frac{1}{2}) = 3 - e$ .
Goal: Find the value of $y (1)$ .
Approach: Rearrange the equation into the standard linear first-order differential equation form $\frac{d y}{d x} + P (x) y = Q (x)$ , determine the integrating factor $C$ , solve via substitution for the RHS integral, determine the constant $C$ using the initial value, and finally evaluate at $x = 1$ .
Execute: Rewrite the equation: $x^{3} \frac{d y}{d x} + x y = 1 ⟹ \frac{d y}{d x} + \frac{1}{x^{2}} y = \frac{1}{x^{3}}$ . Here $P (x) = \frac{1}{x^{2}}$ and $Q (x) = \frac{1}{x^{3}}$ . The integrating factor is $I.F. = e^{\int \frac{1}{x^{2}} d x} = e^{- 1 / x}$ . Multiplying gives: $y e^{- 1 / x} = \int \frac{1}{x^{3}} e^{- 1 / x} d x$ . Substitute $t = - \frac{1}{x}$ , so $d t = \frac{1}{x^{2}} d x$ and $\frac{1}{x} = - t$ : $\int \frac{1}{x^{3}} e^{- 1 / x} d x = \int (- t) e^{t} d t = - (t - 1) e^{t} + C = (1 - t) e^{t} + C = (1 + \frac{1}{x}) e^{- 1 / x} + C$ . Hence, $y (x) = 1 + \frac{1}{x} + C e^{1 / x}$ .
Execute: Use the initial condition $y (\frac{1}{2}) = 3 - e$ : $3 - e = 1 + 2 + C e^{2} = 3 + C e^{2} ⟹ C e^{2} = - e ⟹ C = - e^{- 1} = - \frac{1}{e}$ . Now evaluate at $x = 1$ : $y (1) = 1 + 1 + (- \frac{1}{e}) e^{1} = 2 - 1 = 1$ .
Verify: Check at $x = 1 / 2$ : $y (1 / 2) = 1 + 2 - e^{- 1} \cdot e^{2} = 3 - e$ , which matches the given condition. Also, check $y^{'} (1)$ : with $y (1) = 1$ , $y^{'} = 1 - 1 = 0$ ; from general formula $y^{'} = - 1 / x^{2} - e^{1 / x - 1} / x^{2}$ , at $x = 1$ , $y^{'} (1) = - 1 - (- 1) = 0$ . Solution is completely consistent.

Solving a First-Order Linear Differential Equation Using an Integrating Factor

What feels right?

Rewriting the differential equation in standard linear form yields the integrating factor $e^{- 1 / x}$ , giving $y(x) = rac{1}{x} + 1 - e^{2 - 1/x}$ , which evaluates to $1$ at $x = 1$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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