CUET UG 2025 — Mathematics Calculus

2025mcqeasy

The particular solution of the differential equation $\frac{dy}{dx} = 8yx$ when $y = 1$ at $x = 0$

Official previous-year question

Held on 29 May 2025 · Verified 13 Jul 2026.

Options

A
$y = e^{x^2}$
B
$y = e^{4x^2}$
C
$y = e^{8x^2}$
D
$y = e^x$

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Solution

Given the differential equation $\frac{dy}{dx} = 8yx$ with initial condition $y = 1$ when $x = 0$ .

This is a separable differential equation.

$\frac{dy}{dx} = 8yx$

$\frac{dy}{y} = 8x \, dx$

$\int \frac{dy}{y} = \int 8x \, dx$

$\ln|y| = 4x^2 + C$

$e^{\ln|y|} = e^{4x^2 + C}$

$y = e^{4x^2 + C}$

$y = e^C \cdot e^{4x^2}$

$y = A \cdot e^{4x^2}$ where $A = e^C$

Applying the initial condition $y = 1$ when $x = 0$ :

$1 = A \cdot e^{4(0)^2}$

$1 = A \cdot e^{0}$

$1 = A \cdot 1$

$A = 1$

Therefore, the particular solution is:

$y = e^{4x^2}$

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Solution

Given the differential equation $\frac{dy}{dx} = 8yx$ with initial condition $y = 1$ when $x = 0$ .

This is a separable differential equation.

$\frac{dy}{dx} = 8yx$

$\frac{dy}{y} = 8x \, dx$

$\int \frac{dy}{y} = \int 8x \, dx$

$\ln|y| = 4x^2 + C$

$e^{\ln|y|} = e^{4x^2 + C}$

$y = e^{4x^2 + C}$

$y = e^C \cdot e^{4x^2}$

$y = A \cdot e^{4x^2}$ where $A = e^C$

Applying the initial condition $y = 1$ when $x = 0$ :

$1 = A \cdot e^{4(0)^2}$

$1 = A \cdot e^{0}$

$1 = A \cdot 1$

$A = 1$

Therefore, the particular solution is:

$y = e^{4x^2}$

CUET UG 2025 — Mathematics Calculus

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Solution

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CUET UG 2025 — Mathematics Calculus

Options

Did you get this right?

Solution

Your note

CUET UG Calculus in other years

More CUET UG Calculus questions

Other Mathematics topics for CUET UG

Algebra

Applied-Mathematics

Statistics & Applications

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Calculus