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

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

(x + y − 1) dy = (x + y) dx

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

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

\[\frac{dy}{dx} - y \tan x = e^x \sec x\]

\[\frac{dy}{dx} - y \tan x = e^x\]

(1 + y + x² y) dx + (x + x³) dy = 0

(x² + 1) dy + (2y − 1) dx = 0

`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`

`(2ax+x^2)(dy)/(dx)=a^2+2ax`

(x³ − 2y³) dx + 3x² y dy = 0

x² dy + (x² − xy + y²) dx = 0

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

\[\frac{dy}{dx} + 2y = \sin 3x\]

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

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

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

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

`x cos x(dy)/(dx)+y(x sin x + cos x)=1`

\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]