The equation of the curve that passes through (-1, 1) and the slope of line touching to the curve at any point (x, y) is 3x^2/y, is: Option 1: y^2 = 2x^3 + 3 Option 2: y^3 = 2x^2 + 3 Option 3: y^2 = x^3 + 3/2 Option 4: y^2 = 2x^3 + 6

Answer: y^2 = 2x^3 + 3. Solution: From dy/dx=3x^2/y we get y dy=3x^2 dx, so y^2/2=x^3+c, that is y^2=2x^3+C. Substituting (-1,1) gives 1=-2+C, so C=3 and the curve is y^2=2x^3+3.

Calculus

Differential Equations

Easy

core

The equation of the curve that passes through $(-1, 1)$ and the slope of line touching to the curve at any point $(x, y)$ is $\frac{3x^2}{y}$ , is:

$y^2 = 2x^3 + 3$

$y^3 = 2x^2 + 3$

$y^2 = x^3 + \frac{3}{2}$

$y^2 = 2x^3 + 6$

✅ Correct Option: 1

From $\frac{dy}{dx}=\frac{3x^2}{y}$ we get $y\,dy=3x^2\,dx$ , so $\frac{y^2}{2}=x^3+c$ , that is $y^2=2x^3+C$ . Substituting $(-1,1)$ gives $1=-2+C$ , so $C=3$ and the curve is $y^2=2x^3+3$ .

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

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The particular solution of the differential equation $\left[x \sin^2\left(\frac{y}{x}\right) - y\right]dx + xdy = 0$ , $y = \frac{\pi}{4}$ when $x = 1$ is

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Match the LIST-I with LIST-II

	LIST-I (Differential equation)		LIST-II (Sum of order and degree)
A.	$(y'')^3 + xy' + 2y = 0$	I.	$6$
B.	$(y'')^2 + (y')^3 + 4y = 0$	II.	$3$
C.	$y'' + (y')^2 + 5y = 0$	III.	$4$
D.	$(y'')^4 + 3x(y')^2 + y = 0$	IV.	$5$

Choose the correct answer from the options given below:

2026: 31 May Shift 2

Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Integrating factor
A.	$\dfrac{dy}{dx} + 3y = e^{-2x}$	I.	$\dfrac{1}{x}$
B.	$x\dfrac{dy}{dx} + y = 3x^2$	II.	$e^{-x}$
C.	$x\dfrac{dy}{dx} - y = 3x^2$	III.	$x$
D.	$\dfrac{dy}{dx} - y = x$	IV.	$e^{3x}$

Choose the correct answer from the options given below:

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For $|x| < 1$ , if $x = \cos\left(\frac{1}{a}\log y\right)$ , then

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Solution of differential equation $\frac{dy}{dx} = \frac{1+x^2}{1+y^2}$ is

The equation of the curve that passes through (−1,1)(-1, 1)(−1,1) and the slope of line touching to the curve at any point (x,y)(x, y)(x,y) is 3x2y\frac{3x^2}{y}y3x2​, is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The equation of the curve that passes through $(-1, 1)$ and the slope of line touching to the curve at any point $(x, y)$ is $\frac{3x^2}{y}$ , is: