Commerce (English Medium) इयत्ता १२ - CBSE Question Bank Solutions

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]

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

For the following differential equation, find the general solution:- `y log y dx − x dy = 0`

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sin^{- 1} x\]

For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]

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

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

For the following differential equation, find a particular solution satisfying the given condition:- \[\cos\left( \frac{dy}{dx} \right) = a, y = 1\text{ when }x = 0\]

For the following differential equation, find a particular solution satisfying the given condition:- \[\frac{dy}{dx} = y \tan x, y = 1\text{ when }x = 0\]

Solve the following differential equation:- \[\left( x - y \right)\frac{dy}{dx} = x + 2y\]

Solve the following differential equation:- \[x \cos\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]

Solve the following differential equation:- `y dx + x log  (y)/(x)dy-2x dy=0`

Solve the following differential equation:-

\[\frac{dy}{dx} - y = \cos x\]

Solve the following differential equation:-

\[x\frac{dy}{dx} + 2y = x^2 , x \neq 0\]

Solve the following differential equation:-

\[\frac{dy}{dx} + 2y = \sin x\]

Solve the following differential equation:-

\[\frac{dy}{dx} + 3y = e^{- 2x}\]

Solve the following differential equation:-

\[\frac{dy}{dx} + \frac{y}{x} = x^2\]

Solve the following differential equation:-

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

Solve the following differential equation:-

\[x\frac{dy}{dx} + 2y = x^2 \log x\]

Solve the following differential equation:-

\[x \log x\frac{dy}{dx} + y = \frac{2}{x}\log x\]

Solve the following differential equation:-

(1 + x²) dy + 2xy dx = cot x dx