Calculus

Differential Equations

Easy

common

The product of order and degree of the differential equation

$x\dfrac{d^3y}{dx^3} = \left(1 + \left(\dfrac{dy}{dx}\right)^2\right)^4$ is

✅ Correct Option: 2

The highest derivative is $\dfrac{d^3y}{dx^3}$ , so order $=3$ .

The equation is already polynomial in derivatives and the power of $\dfrac{d^3y}{dx^3}$ is 1, so degree $=1$ .

Product $=3\times 1=3$ .

2022: 16 July Shift 2

Match List I with List II.

List I	List II
A. $\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$	I. order 3, degree 1
B. $\left(\frac{d^2y}{dx^2}\right)^2 = 0$	II. order 2, degree 2
C. $\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + y = 0$	III. order 2, degree 1
D. $\sin\left(\frac{dy}{dx}\right) + 5y = 0$	IV. order 1, degree is not defined

Choose the correct answer from the options given below:

2025: 30 May Shift 1

The product of order and degree of the differential equation $\left(\frac{d^3y}{dx^3}\right)^2 + x^2y\left(\frac{d^2y}{dx^2}\right)^3 = 2x^5$ is:

2026: 21 May Shift 1

The solution of the differential equation $e^{\frac{dy}{dx}} = x+1$ ; $y(0)=5$ is

2026: 22 May Shift 1

General solution of the differential equation $\frac{dy}{dx}=e^{\frac{x^2}{2}}+xy$ is: (where C is an arbitrary constant)

2026: 19 May Shift 1

The solution of the initial value problem $x\frac{dy}{dx} = 2, y(1) = 2$ is

2025: 26 May Shift 2

The solution of the differential equation $xdy - ydx = 0$ represents

The product of order and degree of the differential equation xd3ydx3=(1+(dydx)2)4x\dfrac{d^3y}{dx^3} = \left(1 + \left(\dfrac{dy}{dx}\right)^2\right)^4xdx3d3y​=(1+(dxdy​)2)4 is