General solution of the differential equation x(xdy - ydx) = y dx is :(where c is an arbitrary constant) Option 1: 1/x, = e^((-y)/x+c) Option 2: 1/x, = e^(y/x+c) Option 3: y/x, = e^((-1)/x+c) Option 4: y/x, = e^(1/x+c)

Answer: y/x, = e^((-1)/x+c). Solution: Expanding, x^2dy = y(1+x)dx, so dy/y = (1/x^2+1/x)dx. Integrating gives, y, = -1/x+, x, +c, i.e., y/x, = -1/x+c. Hence, y/x, = e^(-1/x+c).

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General solution of the differential equation $x(xdy - ydx) = y\,dx$ is :(where c is an arbitrary constant)

$\left|\dfrac{1}{x}\right| = e^{\frac{-y}{x}+c}$

$\left|\dfrac{1}{x}\right| = e^{\frac{y}{x}+c}$

$\left|\dfrac{y}{x}\right| = e^{\frac{-1}{x}+c}$

$\left|\dfrac{y}{x}\right| = e^{\frac{1}{x}+c}$

✅ Correct Option: 3

Expanding, $x^2dy = y(1+x)dx$ , so $\dfrac{dy}{y} = \left(\dfrac{1}{x^2}+\dfrac{1}{x}\right)dx$ .

Integrating gives $\log|y| = -\dfrac{1}{x}+\log|x|+c$ , i.e. $\log\left|\dfrac{y}{x}\right| = -\dfrac{1}{x}+c$ .

Hence $\left|\dfrac{y}{x}\right| = e^{-\frac{1}{x}+c}$ .

2026: 14 May Shift 2

Match the LIST-I with LIST-II

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

Choose the correct answer from the options given below:

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The particular solution of the differential equation x(1 + y²)dx - y(1 + x²)dy = 0, y(0) = 1, is:

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The general solution of the differential equation $x\,dx = (x^2+1)\,dy$ is (where C is an arbitrary constant)

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The order of $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \left[a \frac{d^2y}{dx^2}\right]^{\frac{1}{2}}$ is

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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:

General solution of the differential equation x(xdy−ydx)=y dxx(xdy - ydx) = y\,dxx(xdy−ydx)=ydx is :(where c is an arbitrary constant)

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General solution of the differential equation $x(xdy - ydx) = y\,dx$ is :(where c is an arbitrary constant)