Match the LIST-I with LIST-II, LIST-I Differential equations, LIST-II Integrating factor, A., (x+y)dy/dx = 1, I., e^(-x), B., xdy/dx - y = x^2, II., x, C., dy/dx - y = x, III., 1/x, D., xdy/dx + y = x, IV., e^(-y), Choose the correct answer from the options given below: Option 1: A - I, B - III, C - IV, D - II Option 2: A - I, B - II, C - IV, D - III Option 3: A - IV, B - I, C - III, D - II Option 4: A - IV, B - III, C - I, D - II

Answer: A - IV, B - III, C - I, D - II. Solution: A: treat x as the dependent variable, dx/dy-x=y, so IF =e^(-y) (IV). B: dy/dx- yx=x, IF =e^(- x)=1x (III). C: dy/dx-y= x, IF =e^(-x) (I). D: dy/dx+ yx=( x)/x, IF =e^( x)=x (II).

Calculus

Differential Equations

Hard

core

Match the LIST-I with LIST-II

LIST-I
Differential equations LIST-II
Integrating factor
A. $(x+y)\dfrac{dy}{dx} = 1$ I. $e^{-x}$
B. $x\dfrac{dy}{dx} - y = x^2$ II. $x$
C. $\dfrac{dy}{dx} - y = \sin x$ III. $\dfrac{1}{x}$
D. $x\dfrac{dy}{dx} + y = \tan x$ IV. $e^{-y}$

Choose the correct answer from the options given below:

	LIST-I Differential equations		LIST-II Integrating factor
A.	$(x+y)\dfrac{dy}{dx} = 1$	I.	$e^{-x}$
B.	$x\dfrac{dy}{dx} - y = x^2$	II.	$x$
C.	$\dfrac{dy}{dx} - y = \sin x$	III.	$\dfrac{1}{x}$
D.	$x\dfrac{dy}{dx} + y = \tan x$	IV.	$e^{-y}$

A - I, B - III, C - IV, D - II

A - I, B - II, C - IV, D - III

A - IV, B - I, C - III, D - II

A - IV, B - III, C - I, D - II

✅ Correct Option: 4

A: treat $x$ as the dependent variable, $\frac{dx}{dy}-x=y$ , so IF $=e^{-y}$ (IV). B: $\frac{dy}{dx}-\frac yx=x$ , IF $=e^{-\log x}=\frac1x$ (III). C: $\frac{dy}{dx}-y=\sin x$ , IF $=e^{-x}$ (I). D: $\frac{dy}{dx}+\frac yx=\frac{\tan x}{x}$ , IF $=e^{\log x}=x$ (II).

2025: 14 May Shift 2

For the differential equation $(x + y)dy + (x - y)dx = 0$ , which of the following is/are correct?

(A) Differential equation is homogeneous

(B) Order of differential equation is 1

(D) Degree of the equation is not defined

Choose the correct answer from the options given below:

2026: 19 May Shift 1

The general solution of the given differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^{x}+e^{-x}}$ (where C is an arbitrary constant)

2026: 14 May Shift 2

The solution of the differential equation $\frac{dy}{dx} = x^2 + x + \frac{1}{x}$ is

2026: 12 May Shift 2

The general solution of the differential equation $ydx + (x-y^3)dy = 0$ is :

2026: 21 May Shift 2

Match the differential equation (List-I) with its order and degree (List-II)

LIST-I		LIST-II
A. $\dfrac{d^3y}{dx^3} + x^2\left(\dfrac{d^2y}{dx^2}\right) = 0$		I. order =3; degree is not defined
B. $\dfrac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 0$		II. order =3; degree=1
C. $1 - \left(\dfrac{dy}{dx}\right)^2 = \left(a\dfrac{d^2y}{dx^2}\right)^{1/3}$		III. order =2; degree=1
D. $y = x\left(\dfrac{dy}{dx}\right)^3 + \left(\dfrac{d^2y}{dx^2}\right)^4$		IV. order =2; degree=4

Choose the correct answer from the options given below:

2026: 21 May Shift 1

Match List-I with List-II

List-I (Differential Equation)	List-II (Degree)
(A) $2(y'')^2 + 3y = 6$	(I) 4
(B) $2y + 5(y')^4 = 7$	(II) 1
(C) $\log x + (y'')^3 = e^x$	(III) 2
(D) $\sqrt{1-2y'} = 5(y'')^{\frac{1}{4}}$	(IV) 3

Choose the correct answer from the options given below:

Solution

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