What is the solution for y^{'''}-y^'=2sin(t) ?

The solution for y^{'''}-y^'=2sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + cos (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + cos (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2} t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} sin (2 x)

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

y^{^{''}} + 16 y^{^{'}} + 64 y = t^{- 2} e^{- 8 t}

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 4 t

What is the solution for y^{'''}-y^'=2sin(t) ?
The solution for y^{'''}-y^'=2sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+cos(t)