Why substitute rather than ?

Either works, but choosing gives , keeping standard positive signs in the linear equation.

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Differential Equations: Mathematics | JEE Main

Let

y = y(x)

be the solution curve of the differential equation

\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log_e x)), x > 0, y(1) = 3 \text{. Then } \frac{y^2(x)}{9} \text{ is equal to :}

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What feels right?

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

\frac{x^2}{2x^3(2+\log_e x^3)-3}

\frac{x^2}{3x^3(1+\log_e x^2)-2}

\frac{x^2}{7-3x^3(2+\log_e x^2)}

\frac{x^2}{5-2x^3(2+\log_e x^3)}

Step-by-step solutionView

Correct answer

The expression for $\frac{y^2(x)}{9}$ is $\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$ .

Option analysis

Why each option works or fails

A ·

\frac{x^2}{2x^3(2+\log_e x^3)-3}

Evaluating the denominator at $x = 1$ gives $2(1)(2) - 3 = 1$ , which satisfies the initial condition but arises from an incorrect integration constant sign or erroneous algebraic rearrangement of the Bernoulli equation. Carry out the substitution $v = y^{-2}$ carefully: $\frac{dv}{dx} + \frac{2}{x}v = -2(1 + \log_e x)$ , ensuring the negative sign from the derivative is properly applied to the non-homogeneous term.

B ·

\frac{x^2}{3x^3(1+\log_e x^2)-2}

Evaluating at $x = 1$ yields $3(1)(1) - 2 = 1$ , matching the initial condition but resulting from incorrectly integrating by parts or misapplying the power rule on logarithmic terms. When computing $\int 2x^2(1 + \log_e x) dx$ , integrate $x^2 \log_e x$ accurately to get $\frac{x^3}{3}\log_e x - \frac{x^3}{9}$ , leading to a factor of $x^3(2 + 3\log_e x)$ rather than $(1 + \log_e x^2)$ .

C ·

\frac{x^2}{7-3x^3(2+\log_e x^2)}

Evaluating at $x = 1$ yields $7 - 3(1)(2) = 1$ , which matches the initial value but contains an arithmetic error in distributing the integrating factor constant across terms. Ensure the coefficient of $x^3$ matches the integrated result $-\frac{2}{3}x^3 - 2(\frac{x^3}{3}\log_e x - \frac{x^3}{9}) = -\frac{2}{9}x^3(4 + 3\log_e x) = -\frac{2}{9}x^3(4 + \log_e x^3)$ .

D ·

\frac{x^2}{5-2x^3(2+\log_e x^3)}

This is the correct option. Testing at $x = 1$ gives $\frac{1}{5 - 2(2)} = 1 = \frac{3^2}{9}$ , and solving the Bernoulli equation via $v = y^{-2}$ yields the exact function. None required.

Reviewed route

Solution

StepWorking

01given

The differential equation is $\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \ln x))$ for $x > 0$ , with initial condition $y(1) = 3$ .

02approach

Rearrange the equation to the Bernoulli form $\frac{dy}{dx} - \frac{1}{x}y = y^3(1 + \ln x)$ . Divide by $y^3$ and substitute $t = -\frac{1}{y^2}$ to reduce it to a first-order linear differential equation in $t$ .

03execute

Dividing by $y^3$ gives $y^{-3}\frac{dy}{dx} - \frac{1}{x}y^{-2} = 1 + \ln x$ . Let $t = -y^{-2}$ , so $\frac{dt}{dx} = 2y^{-3}\frac{dy}{dx}$ , or $y^{-3}\frac{dy}{dx} = \frac{1}{2}\frac{dt}{dx}$ . Substituting yields $\frac{1}{2}\frac{dt}{dx} + \frac{t}{x} = 1 + \ln x \implies \frac{dt}{dx} + \frac{2}{x}t = 2(1 + \ln x)$ .

04execute

Find the integrating factor: $I.F. = e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2$ . The general solution for $t$ is $t \cdot x^2 = \int 2x^2(1 + \ln x)dx = 2\int x^2 dx + 2\int x^2 \ln x dx$ . Using integration by parts for the second integral: $\int x^2 \ln x dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}dx = \frac{x^3}{3}\ln x - \frac{x^3}{9}$ . Thus, $t x^2 = \frac{2x^3}{3} + \frac{2x^3}{3}\ln x - \frac{2x^3}{9} + C = \frac{4x^3}{9} + \frac{2x^3}{3}\ln x + C$ .

05execute

Substitute $t = -\frac{1}{y^2}$ : $-\frac{x^2}{y^2} = \frac{2x^3}{9}(2 + 3\ln x) + C$ . Using the initial condition $x = 1, y = 3$ : $-\frac{1^2}{3^2} = \frac{2(1)^3}{9}(2 + 0) + C \implies -\frac{1}{9} = \frac{4}{9} + C \implies C = -\frac{5}{9}$ .

06execute

Rewrite the equation: $-\frac{x^2}{y^2} = \frac{2x^3(2 + \ln x^3) - 5}{9} \implies \frac{x^2}{y^2} = \frac{5 - 2x^3(2 + \ln x^3)}{9} \implies \frac{y^2(x)}{9} = \frac{x^2}{5 - 2x^3(2 + \ln x^3)}$ .

✓verify

At $x = 1$ : $\frac{y^2(1)}{9} = \frac{1^2}{5 - 2(1)(2 + 0)} = \frac{1}{5 - 4} = 1 \implies y(1)^2 = 9 \implies y(1) = 3$ , which matches the given initial condition.

✓ 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 substitute

t = -1/y^2

rather than

t = 1/y^2

Either works, but choosing $t = -1/y^2$ gives $\frac{dt}{dx} = +\frac{2}{y^3}\frac{dy}{dx}$ , keeping standard positive signs in the linear equation.

Answer

The expression for $\frac{y^2(x)}{9}$ is $\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$ .

Why each option works or fails

A: $\frac{x^2}{2x^3(2+\log_e x^3)-3}$ - Evaluating the denominator at $x = 1$ gives $2(1)(2) - 3 = 1$ , which satisfies the initial condition but arises from an incorrect integration constant sign or erroneous algebraic rearrangement of the Bernoulli equation. Carry out the substitution $v = y^{-2}$ carefully: $\frac{dv}{dx} + \frac{2}{x}v = -2(1 + \log_e x)$ , ensuring the negative sign from the derivative is properly applied to the non-homogeneous term.
B: $\frac{x^2}{3x^3(1+\log_e x^2)-2}$ - Evaluating at $x = 1$ yields $3(1)(1) - 2 = 1$ , matching the initial condition but resulting from incorrectly integrating by parts or misapplying the power rule on logarithmic terms. When computing $\int 2x^2(1 + \log_e x) dx$ , integrate $x^2 \log_e x$ accurately to get $\frac{x^3}{3}\log_e x - \frac{x^3}{9}$ , leading to a factor of $x^3(2 + 3\log_e x)$ rather than $(1 + \log_e x^2)$ .
C: $\frac{x^2}{7-3x^3(2+\log_e x^2)}$ - Evaluating at $x = 1$ yields $7 - 3(1)(2) = 1$ , which matches the initial value but contains an arithmetic error in distributing the integrating factor constant across terms. Ensure the coefficient of $x^3$ matches the integrated result $-\frac{2}{3}x^3 - 2(\frac{x^3}{3}\log_e x - \frac{x^3}{9}) = -\frac{2}{9}x^3(4 + 3\log_e x) = -\frac{2}{9}x^3(4 + \log_e x^3)$ .
D · correct: $\frac{x^2}{5-2x^3(2+\log_e x^3)}$ - This is the correct option. Testing at $x = 1$ gives $\frac{1}{5 - 2(2)} = 1 = \frac{3^2}{9}$ , and solving the Bernoulli equation via $v = y^{-2}$ yields the exact function. None required.

Step-by-step solution

given: The differential equation is $\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \ln x))$ for $x > 0$ , with initial condition $y(1) = 3$ .
approach: Rearrange the equation to the Bernoulli form $\frac{dy}{dx} - \frac{1}{x}y = y^3(1 + \ln x)$ . Divide by $y^3$ and substitute $t = -\frac{1}{y^2}$ to reduce it to a first-order linear differential equation in $t$ .
execute: Dividing by $y^3$ gives $y^{-3}\frac{dy}{dx} - \frac{1}{x}y^{-2} = 1 + \ln x$ . Let $t = -y^{-2}$ , so $\frac{dt}{dx} = 2y^{-3}\frac{dy}{dx}$ , or $y^{-3}\frac{dy}{dx} = \frac{1}{2}\frac{dt}{dx}$ . Substituting yields $\frac{1}{2}\frac{dt}{dx} + \frac{t}{x} = 1 + \ln x \implies \frac{dt}{dx} + \frac{2}{x}t = 2(1 + \ln x)$ .
execute: Find the integrating factor: $I.F. = e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2$ . The general solution for $t$ is $t \cdot x^2 = \int 2x^2(1 + \ln x)dx = 2\int x^2 dx + 2\int x^2 \ln x dx$ . Using integration by parts for the second integral: $\int x^2 \ln x dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}dx = \frac{x^3}{3}\ln x - \frac{x^3}{9}$ . Thus, $t x^2 = \frac{2x^3}{3} + \frac{2x^3}{3}\ln x - \frac{2x^3}{9} + C = \frac{4x^3}{9} + \frac{2x^3}{3}\ln x + C$ .
execute: Substitute $t = -\frac{1}{y^2}$ : $-\frac{x^2}{y^2} = \frac{2x^3}{9}(2 + 3\ln x) + C$ . Using the initial condition $x = 1, y = 3$ : $-\frac{1^2}{3^2} = \frac{2(1)^3}{9}(2 + 0) + C \implies -\frac{1}{9} = \frac{4}{9} + C \implies C = -\frac{5}{9}$ .
execute: Rewrite the equation: $-\frac{x^2}{y^2} = \frac{2x^3(2 + \ln x^3) - 5}{9} \implies \frac{x^2}{y^2} = \frac{5 - 2x^3(2 + \ln x^3)}{9} \implies \frac{y^2(x)}{9} = \frac{x^2}{5 - 2x^3(2 + \ln x^3)}$ .
verify: At $x = 1$ : $\frac{y^2(1)}{9} = \frac{1^2}{5 - 2(1)(2 + 0)} = \frac{1}{5 - 4} = 1 \implies y(1)^2 = 9 \implies y(1) = 3$ , which matches the given initial condition.

Shortcut:

When to use it: Use when the question asks for an explicit function and provides an initial condition $y(x_0) = y_0$ .

concept: The solution must satisfy the initial condition $x = 1 \implies y = 3$ , meaning $\frac{y^2(1)}{9} = \frac{3^2}{9} = 1$ .

option_verdict: At $x = 1$ , the expression evaluates to $\frac{1}{2(1)(2 + 0) - 3} = \frac{1}{4 - 3} = 1$ . While it matches at $x=1$ , differentiation shows the sign of the constant leads to an incorrect derivative sign.

option_verdict: At $x = 1$ , the expression evaluates to $\frac{1}{3(1)(1 + 0) - 2} = \frac{1}{3 - 2} = 1$ , but log term has $x^2$ instead of $x^3$ derived from $\int x^2 \ln x dx$ .

option_verdict: At $x = 1$ , the denominator evaluates to $7 - 3(2 + 0) = 1$ , yielding 1, but the logarithmic power is $x^2$ instead of $x^3$ .

option_verdict: At $x = 1$ , the expression evaluates to $\frac{1}{5 - 2(1)(2 + 0)} = \frac{1}{5 - 4} = 1$ , and matches the exact integration coefficients.

discriminator: In $\int 2x^2 \ln x dx$ , the term $\frac{2x^3}{3}\ln x = \frac{2x^3}{9}\ln(x^3)$ requires the logarithmic argument inside the bracket to be $x^3$ , immediately eliminating options (1) and (2).

Differential Equations: Mathematics | JEE Main

What feels right?

The expression for $\frac{y^2(x)}{9}$ is $\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

06execute

✓verify

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