What is the solution for y^{''}-y^'-2y=10,y(-1)=0,y^'(-1)=1 ?

The solution for y^{''}-y^'-2y=10,y(-1)=0,y^'(-1)=1 is y=2e^{2+2t}+3e^{-t-1}-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-2y=10,y(-1)=0,y^'(-1)=1

y^{^{''}} - y^{^{'}} - 2 y = 10, y (- 1) = 0, y^{^{'}} (- 1) = 1

y = 2 e^{2 + 2 t} + 3 e^{- t - 1} - 5

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 10

Solve linear ODE:

y = 2 e^{2 + 2 t} + 3 e^{- t - 1} - 5

y = 2 e^{2 + 2 t} + 3 e^{- t - 1} - 5

y^{^{''}} - 3 y^{^{'}} + 2 y = 1

y^{^{''}} + 36 y = 288 sin (6 x) + 96 cos (6 x)

y^{^{''}} - y^{^{'}} - 2 y = 1, + t y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 10 sin (x)

(D^{2} + 2 D + 2) y = x^{2} + x + 1

What is the solution for y^{''}-y^'-2y=10,y(-1)=0,y^'(-1)=1 ?
The solution for y^{''}-y^'-2y=10,y(-1)=0,y^'(-1)=1 is y=2e^{2+2t}+3e^{-t-1}-5