en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+2y-5=0

Solution

y^{^{'}} + 2 y - 5 = 0

Solution

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

Solution steps

y^{'} (t) + 2 y - 5 = 0

Solve separable ODE:

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

y = \frac{e ^{- 2 t - c_{1}}}{2} + \frac{5}{2}

Popular Examples

(dy)/(dx)y=(6x^2+4x^2)/(3y^6)

\frac{d y}{d x} y = \frac{6 x ^{2} + 4 x ^{2}}{3 y ^{6}}

y^{^{'}} = 3 x y + 5 y

y^'=(cos^2(x))/(y^2)

y^{^{'}} = \frac{cos ^{2} ( x )}{y ^{2}}

y^2y^'=x^2+6e^x

y^{2} y^{^{'}} = x^{2} + 6 e^{x}

(dy)/(dx)=((3x^2+4x+2))/(2(y-1))

\frac{d y}{d x} = \frac{( 3 x ^{2} + 4 x + 2 )}{2 ( y - 1 )}

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y-5=0 ?
The solution for y^'+2y-5=0 is y=(e^{-2t-c_{1}})/2+5/2