The solution of the differential equation dy/dx + y = 1, y(1) = 0 is Option 1: 1+x = _e 1-y Option 2: 1-x = _e 1+y Option 3: 1+x = _e 1+y Option 4: 1-x = _e 1-y

Answer: 1-x = _e 1-y. Solution: Separating, dy/(1-y)=dx gives -, 1-y, =x+C. Using y(1)=0: 0=1+C, so C=-1 and, 1-y, =1-x.

Calculus

Differential Equations

Medium

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The solution of the differential equation $\frac{dy}{dx} + y = 1, y(1) = 0$ is

$1+x = \log_e|1-y|$

$1-x = \log_e|1+y|$

$1+x = \log_e|1+y|$

$1-x = \log_e|1-y|$

✅ Correct Option: 4

Separating, $\frac{dy}{1-y}=dx$ gives $-\log|1-y|=x+C$ . Using $y(1)=0$ : $0=1+C$ , so $C=-1$ and $\log|1-y|=1-x$ .

2026: 29 May Shift 1

The general solution of the differential equation $(x+1)\dfrac{dy}{dx}=2e^{-y}-1$ is:

(where C is an arbitrary constant)

2026: 21 May Shift 1

For the differential equation $\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} = k, k \neq 0$ , which one of the following is TRUE ?

2025: 22 May Shift 1

The solution of the differential equation $\frac{dr}{dt} = -rt, r(0) = r_0$ is

2022: 30 Aug Shift 1

Order and degree of the differential equation $y\frac{dy}{dx} + \frac{4}{\frac{dy}{dx}} = 5$ are

2026: 15 May Shift 1

If $c$ is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I (Differential Equation)		LIST-II (Solution)
A.	$\dfrac{dy}{dx} = 1 + x + y + xy$	I.	$\log_e \vert y \vert = 2x + \log_e (x-1)^2 + c$
B.	$(x-1)\dfrac{dy}{dx} = 2xy$	II.	$y = cx$
C.	$\dfrac{dy}{dx} = 1 - x + y - xy$	III.	$\log_e \vert 1+y \vert = x + \dfrac{x^2}{2} + c$
D.	$x\,dy = y\,dx$	IV.	$\log_e \vert 1+y \vert = x - \dfrac{x^2}{2} + c$

Choose the correct answer from the options given below:

2022: 23 Aug Shift 1

Match List - I with List - II

List - I (Differential Equation)	List - II (Degree)
A. $\left[1 + (y')^2\right]^2 = y''$	I. $2$
B. $\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3$	II. $4$
C. $(y''')^2 + y'' + 3y' + 5y = e^x$	III. $1$
D. $\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2$	IV. $3$

Choose the correct answer from the option given below:

The solution of the differential equation dydx+y=1,y(1)=0\frac{dy}{dx} + y = 1, y(1) = 0dxdy​+y=1,y(1)=0 is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The solution of the differential equation $\frac{dy}{dx} + y = 1, y(1) = 0$ is