en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-8=0

Solution

y^{^{''}} + 2 y^{^{'}} - 8 = 0

Solution

y = c_{1} + c_{2} e^{- 2 t} + 4 t

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + 4 t

y = c_{1} + c_{2} e^{- 2 t} + 4 t

Graph

Popular Examples

0.5q^{''}+20q^'+1/(0.0025)q=100

0.5 q^{^{''}} + 20 q^{^{'}} + \frac{1}{0.0025} q = 100

x^{''}+16x=14cos(3t)

x^{^{''}} + 16 x = 14 cos (3 t)

y^{''}-2y^'+y=t*e^t

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t}

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

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

(D^2-D-2)y=6e^{(-t)/2}

(D^{2} - D - 2) y = 6 e^{\frac{- t}{2}}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+2y^'-8=0 ?
The solution for y^{''}+2y^'-8=0 is y=c_{1}+c_{2}e^{-2t}+4t