The particular solution of the differential equation (dy/dx)= 3x + 4y satisfying y = 0 when x = 0 is: Option 1: 3e³ˣ + 4e⁻⁴ʸ + 7 = 0 Option 2: 3e³ˣ - 4e⁻⁴ʸ - 7 = 0 Option 3: 4e³ˣ + 3e⁻⁴ʸ - 7 = 0 Option 4: 4e³ˣ - 3e⁻⁴ʸ + 7 = 0

Answer: 4e³ˣ + 3e⁻⁴ʸ - 7 = 0. Solution: The differential equation is (dy/dx) = 3x + 4y with condition y = 0 when x = 0. In differential equations, typically means natural logarithm. Applying exponential to both sides: dy/dx = e^(3x + 4y) dy/dx = e^3x * e^4y Dividing both sides by e^4y: 1/e^4y * dy = e^3x * dx e^(-4y) dy = e^3x dx Integrating both sides: e^(-4y) dy = e^3x dx -(e^(-4y))/4 = e^3x/3 + C Applying the initial condition when x = 0 and y = 0: -e^0/4 = e^0/3 + C -1/4 = 1/3 + C C = -1/4 - 1/3 C = -3/12 - 4/12 C = -7/12 Substituting C = -7/12: -(e^(-4y))/4 = e^3x/3 - 7/12 Multiplying everything by 12: -3e^(-4y) = 4e^3x - 7 3e^(-4y) = -4e^3x + 7 4e^3x + 3e^(-4y) - 7 = 0 Therefore, the particular solution is 4e^3x + 3e^(-4y) - 7 = 0

Calculus

Differential Equations

Medium

common

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right)= 3x + 4y$ satisfying $y = 0$ when $x = 0$ is:

$3e³ˣ + 4e⁻⁴ʸ + 7 = 0$

$3e³ˣ - 4e⁻⁴ʸ - 7 = 0$

$4e³ˣ + 3e⁻⁴ʸ - 7 = 0$

$4e³ˣ - 3e⁻⁴ʸ + 7 = 0$

✅ Correct Option: 3

The differential equation is $\log\left(\frac{dy}{dx}\right) = 3x + 4y$ with condition $y = 0$ when $x = 0$ .

In differential equations, $\log$ typically means natural logarithm.

Applying exponential to both sides:

$\frac{dy}{dx} = e^{3x + 4y}$

$\frac{dy}{dx} = e^{3x} \cdot e^{4y}$

Dividing both sides by $e^{4y}$ :

$\frac{1}{e^{4y}} \cdot dy = e^{3x} \cdot dx$

$e^{-4y} \, dy = e^{3x} \, dx$

Integrating both sides:

$\int e^{-4y} \, dy = \int e^{3x} \, dx$

$-\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} + C$

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

$-\frac{e^{0}}{4} = \frac{e^{0}}{3} + C$

$-\frac{1}{4} = \frac{1}{3} + C$

$C = -\frac{1}{4} - \frac{1}{3}$

$C = -\frac{3}{12} - \frac{4}{12}$

$C = -\frac{7}{12}$

Substituting $C = -\frac{7}{12}$ :

$-\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} - \frac{7}{12}$

Multiplying everything by 12:

$-3e^{-4y} = 4e^{3x} - 7$

$3e^{-4y} = -4e^{3x} + 7$

$4e^{3x} + 3e^{-4y} - 7 = 0$

Therefore, the particular solution is $4e^{3x} + 3e^{-4y} - 7 = 0$

2022: 6 Aug Shift 2

If the solution curve of the differentiable equation $\frac{dy}{dx} + 2y = e^{3x}$ , passes through the point $\left(0, \frac{6}{5}\right)$ , then the value of $y(\log_e 2)$ is:

2026: 12 May Shift 2

Match List-I with List-II

List-I (Differential Equation)	List-II (Order and degree)
(A) $\left(\dfrac{d^2y}{dx^2}\right)^2 + e^{\frac{dy}{dx}} = 0$	(I) order = 2, degree = 1
(B) $xy\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 - y\dfrac{dy}{dx} = 0$	(II) order = 2, degree = not defined
(C) $\left(\dfrac{d^3y}{dx^3}\right)^2 + \log_e\left(\dfrac{dy}{dx}\right) = 4x$	(III) order = 3, degree = 2
(D) $\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^2}\right) + y\dfrac{dy}{dx} = \left(1+\dfrac{dy}{dx}\right)^{1/2}$	(IV) order = 3, degree = not defined

Choose the correct answer from the options given below:

2025: 21 May Shift 1

The number of arbitrary constants in the general solution of a differential equation with degree 1 and order 3, is

2025: 30 May Shift 1

The differential equation representing the family of curves $y = Ax + \frac{B}{x}$ , $x \neq 0$ where A and B are arbitrary constants, is given by

2026: 31 May Shift 2

Match the LIST-I with LIST-II

[ $a$ is an arbitrary constant]

	LIST-I Differential Equation		LIST-II Family of curves
A.	$x\,dy - y\,dx = 0$	I.	$x^2 + y^2 = a^2$
B.	$2y\,dx - x\,dy = 0$	II.	$y = ax$
C.	$x\,dx + y\,dy = 0$	III.	$y^2 = 16x + a$
D.	$y\,dy - 8\,dx = 0$	IV.	$x^2 = ay$

Choose the correct answer from the options given below:

2025: 14 May Shift 1

Consider the differential equation, $x\frac{dy}{dx} = y(\log_e y - \log_e x + 1)$ , then which of the following are true?

(A) It is a linear differential equation

(B) It is a homogenous differential equation

(D) Its general solution is $\log_e\left(\frac{x}{y}\right) = Cy$ , where C is constant of integration

(E) If $y(1) = 1$ , then its particular solution is $y = x$

Choose the correct answer from the options given below:

The particular solution of the differential equation log⁡(dydx)=3x+4y\log\left(\frac{dy}{dx}\right)= 3x + 4ylog(dxdy​)=3x+4y satisfying y=0y = 0y=0 when x=0x = 0x=0 is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right)= 3x + 4y$ satisfying $y = 0$ when $x = 0$ is: