For the Following Differential Equation, Find the General Solution:- D Y D X = ( 1 + X 2 ) ( 1 + Y 2 )

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]

योग

We have,

\[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]

\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = \left( 1 + x^2 \right)dx\]

Integrating both sides, we get

\[\int\frac{1}{\left( 1 + y^2 \right)}dy = \int\left( 1 + x^2 \right)dx\]

\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]

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अध्याय 21: Differential Equations - Revision Exercise [पृष्ठ १४६]

अध्याय 21 Differential Equations
Revision Exercise | Q 64.3 | पृष्ठ १४६

The solution of the differential equation dy/dx = sec x – y tan x is:

(A) y sec x = tan x + c

(B) y sec x + tan x = c

(D) sec x + y tan x = c

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`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x : (y sin y + cos y + x) y′ = y

The number of arbitrary constants in the particular solution of a differential equation of third order are ______.

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\[\frac{dy}{dx} = \left( x + y \right)^2\]

\[\frac{dy}{dx} + 5y = \cos 4x\]

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For the following differential equation, find a particular solution satisfying the given condition:

\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]