Solution of differential equation dy/dx = (1+x^2)/(1+y^2) is Option 1: x^3 - y^3 + 3(x+y) + C = 0 : where C is an arbitrary constant Option 2: x^3 + y^3 + 3(x-y) + C = 0: where C is an arbitrary constant Option 3: x^3 - y^3 + 3(x-y) + C = 0: where C is an arbitrary constant Option 4: x^3 + y^3 + 3(x+y) + C = 0: where C is an arbitrary constant

Answer: x^3 - y^3 + 3(x-y) + C = 0: where C is an arbitrary constant. Solution: Separate the variables: (1+y^2)dy=(1+x^2)dx. Integrating gives y+y^3/3=x+x^3/3+c. Multiply by 3 and shift all terms to one side: x^3-y^3+3x-3y+C=0, i.e. x^3-y^3+3(x-y)+C=0.

Calculus

Differential Equations

Easy

common

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

$x^3 - y^3 + 3(x+y) + C = 0$ : where $C$ is an arbitrary constant

$x^3 + y^3 + 3(x-y) + C = 0$ : where $C$ is an arbitrary constant

$x^3 - y^3 + 3(x-y) + C = 0$ : where $C$ is an arbitrary constant

$x^3 + y^3 + 3(x+y) + C = 0$ : where $C$ is an arbitrary constant

✅ Correct Option: 3

Separate the variables: $(1+y^2)dy=(1+x^2)dx$ . Integrating gives $y+\frac{y^3}{3}=x+\frac{x^3}{3}+c$ . Multiply by 3 and shift all terms to one side: $x^3-y^3+3x-3y+C=0$ , i.e. $x^3-y^3+3(x-y)+C=0$ .

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The order and the degree of the differential equation $\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = 2x^2 \log\left(\frac{d^2y}{dx^2}\right)$ are respectively:

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

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The particular solution of the differential equation $\frac{dy}{dx}=x(2\log x+1)$ , given that $y=0$ when $x=2$ , is:

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

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If $U(x) = x + \sqrt{1 + x^2}$ , then solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0$ , is :

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The differential equation representing family of curves $y = ae^{mx} + be^{nx}$ , where $a$ and $b$ are arbitrary constants, is

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Match List-I with List-II

List-I	List-II
Differential equation	Order and degree
(A) $(y'')^3 + (y')^4 - 6 = (y''')^2$	(I) Order = 1, Degree = 2
(B) $\sqrt{(y')^2 + 5} = y''$	(II) Order = 2, Degree = 3
(C) $(y')^2 = (2 + y'')^{3/2}$	(III) Order = 2, Degree = 2
(D) $y = xy' + \sqrt{a^2(y')^2 + b^2}$	(IV) Order = 3, Degree = 2

Choose the correct answer from the options given below:

Solution of differential equation dydx=1+x21+y2\frac{dy}{dx} = \frac{1+x^2}{1+y^2}dxdy​=1+y21+x2​ is

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