Solution of dy/dx = (1+x^2)(1+y^2) is: Option 1: ^-1 y = x + x^3/3 + c Option 2: ^-1 y = x - x^3/3 + c Option 3: ^-1 y = x^2 + x^3/3 + c Option 4: ^-1 y = x^2 - x^3/3 + c

The answer is '^-1 y = x + x^3/3 + c'. View full solution:

CUET Mathematics - Solution of dy/dx = (1+x^2)(1+y^2) is: | PYQs + Solutions

CUET Mathematics 2022 25 May Shift 1

Calculus

Easy

$\tan^{-1} y = x + \frac{x^3}{3} + c$

$\tan^{-1} y = x - \frac{x^3}{3} + c$

$\tan^{-1} y = x^2 + \frac{x^3}{3} + c$

$\tan^{-1} y = x^2 - \frac{x^3}{3} + c$

✅ Correct Option: 1

Related questions:

30 May Shift 1

Which of the following are first order linear differential equations?

(A) $\frac{dx}{dy} + P_1(y)x = Q_1(y)$ : $P_1(y)$ and $Q_1(y)$ are functions of y or constant functions

(B) $\frac{dy}{dx} + P_2(x)y = Q_2(x)$ : $P_2(x)$ and $Q_2(x)$ are functions of x or constant functions

(D) $(1 + x^2)\frac{dy}{dx} - 2xy = x^2 + 3$

Choose the correct answer from the options given below:

4 Aug Shift 1

The solution of the differential equation $\frac{dy}{dx} = \frac{\lambda^2}{(x+y)^2}$ ( $\lambda$ is constant) is:

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: