The particular solution of the differential equation dy/dx = 8yx when y = 1 at x = 0 Option 1: y = e^x^2 Option 2: y = e^4x^2 Option 3: y = e^8x^2 Option 4: y = e^x

Answer: y = e^4x^2. Solution: Given the differential equation dy/dx = 8yx with initial condition y = 1 when x = 0. This is a separable differential equation. dy/dx = 8yx dy/y = 8x dx dy/y = 8x dx, y, = 4x^2 + C e^(, y, ) = e^(4x^2 + C) y = e^(4x^2 + C) y = e^C * e^4x^2 y = A * e^4x^2 where A = e^C Applying the initial condition y = 1 when x = 0: 1 = A * e^(4(0)^2) 1 = A * e^0 1 = A * 1 A = 1 Therefore, the particular solution is: y = e^4x^2

Calculus

Differential Equations

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

$y = e^{x^2}$

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

$y = e^{8x^2}$

$y = e^x$

✅ Correct Option: 2

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:

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

2025: 15 May Shift 1

Consider the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ , then which of the following statements are correct?

(A) It is homogeneous

(B) It has $\sec x$ as its integrating factor

(D) It's degree is not defined

Choose the correct answer from the options given below:

2023: 22 May Shift 3

The differential equation $y = xp + \sqrt{x^2 p^3 + 4}$ where $p = \frac{dy}{dx}$ is :

(A) of order 1

(B) of degree 1

(D) of degree 3

Choose the correct answer from the options given below :

2023: 15 June Shift 2

The differential equation whose solution is $Ax^2 + By^2 = 1$ where A and B are arbitrary constant is of :

(A) first order and first degree

(B) second order and first degree

(D) second order

Choose the correct answer from the options given below :

2026: 18 May Shift 1

Solution of differential equation $\frac{dy}{dx} = e^{x+y} - xe^y$ is

2022: 7 Aug Shift 2

If curve represented by differential equation $x \frac{dy}{dx} + y = e^x$ passes through (1, 1), then $y(-1)$ is :

2026: 19 May Shift 2

The general solution of the given differential equation $dy + (2y - e^{-2x})dx = 0$ , is:

[where c is an arbitrary constant]

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

Solution

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