What is the solution for y^{''}+y^'-2y=4,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+y^'-2y=4,y(0)=2,y^'(0)=1 is y=3e^t+e^{-2t}-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'-2y=4,y(0)=2,y^'(0)=1

y^{^{''}} + y^{^{'}} - 2 y = 4, y (0) = 2, y^{^{'}} (0) = 1

y = 3 e^{t} + e^{- 2 t} - 2

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 4

Solve linear ODE:

y = 3 e^{t} + e^{- 2 t} - 2

y = 3 e^{t} + e^{- 2 t} - 2

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

y^{^{''}} + y = sin (3 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 1 + x

y^{^{''}} - 8 y^{^{'}} + 16 y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} = sin (4 x)

What is the solution for y^{''}+y^'-2y=4,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+y^'-2y=4,y(0)=2,y^'(0)=1 is y=3e^t+e^{-2t}-2