The general solution of the differential equation dy/dx-y/x+cosec(y/x)=0 is (Consider x=_e x) Option 1: (x/y)= x+C, where C is an arbitrary constant Option 2: (y/x)= x+C, where C is an arbitrary constant Option 3: (y/x)= x+C, where C is an arbitrary constant Option 4: (x/y)= x+C, where C is an arbitrary constant

Answer: (y/x)= x+C, where C is an arbitrary constant. Solution: This is homogeneous, so put y=vx: v+xdv/dx=v-cosec v, giving v dv=-dx/x. Integrating, - v=- x+C', i.e. (y/x)= x+C.

Calculus

Differential Equations

Medium

core

The general solution of the differential equation $\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left(\frac{y}{x}\right)=0$ is

(Consider $\log x=\log_e x$ )

$\sin\left(\frac{x}{y}\right)=\log x+C$ , where C is an arbitrary constant

$\sin\left(\frac{y}{x}\right)=\log x+C$ , where C is an arbitrary constant

$\cos\left(\frac{y}{x}\right)=\log x+C$ , where C is an arbitrary constant

$\cos\left(\frac{x}{y}\right)=\log x+C$ , where C is an arbitrary constant

✅ Correct Option: 3

This is homogeneous, so put $y=vx$ : $v+x\frac{dv}{dx}=v-\text{cosec}\,v$ , giving $\sin v\,dv=-\frac{dx}{x}$ . Integrating, $-\cos v=-\log x+C'$ , i.e. $\cos\left(\frac{y}{x}\right)=\log x+C$ .

2025: 15 May Shift 1

Match List-I with List-II

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

Choose the correct answer from the options given below:

2025: 2 June Shift 1

The solution of the differential equation $ydx + (x - y^2)dy = 0$ is

2026: 31 May Shift 2

Solution of differential equation $x\,dy - y\,dx = 0$ represents:

2026: 22 May Shift 2

Let y(x) be the solution curve of the differential equation $x(1+y^2)dx - y(1+x^2)dy = 0$ and it passes through the point (0, 1), then

2026: 22 May Shift 1

The general solution of the differential equation $\frac{dy}{dx}=e^{2x-y}+xe^{-y}$ is:

2026: 29 May Shift 2

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

The general solution of the differential equation dydx−yx+cosec(yx)=0\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left(\frac{y}{x}\right)=0dxdy​−xy​+cosec(xy​)=0 is (Consider log⁡x=log⁡ex\log x=\log_e xlogx=loge​x)

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CUET Mathematics Past Year Questions (Topic-Wise)

The general solution of the differential equation $\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left(\frac{y}{x}\right)=0$ is

(Consider $\log x=\log_e x$ )