What is the solution for y^{''}+y^'-2y=2t-40cos(2t) ?

The solution for y^{''}+y^'-2y=2t-40cos(2t) is y=c_{1}e^t+c_{2}e^{-2t}-t-1/2-2sin(2t)+6cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'-2y=2t-40cos(2t)

y^{^{''}} + y^{^{'}} - 2 y = 2 t - 40 cos (2 t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2} - 2 sin (2 t) + 6 cos (2 t)

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 2 t - 40 cos (2 t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2} - 2 sin (2 t) + 6 cos (2 t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2} - 2 sin (2 t) + 6 cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 45 y = - 4 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{3} t

y^{^{''}} + 16 y = sin (x)

x^{^{''}} + 36 x = 100 cos (4 t)

y^{^{''}} - 5 y^{^{'}} = 8 e^{4 t} - 4 e^{- t}

What is the solution for y^{''}+y^'-2y=2t-40cos(2t) ?
The solution for y^{''}+y^'-2y=2t-40cos(2t) is y=c_{1}e^t+c_{2}e^{-2t}-t-1/2-2sin(2t)+6cos(2t)