What is the solution for x^{''}+4x=2e^t ?

The solution for x^{''}+4x=2e^t is x=c_{1}cos(2t)+c_{2}sin(2t)+2/5 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+4x=2e^t

x^{^{''}} + 4 x = 2 e^{t}

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{2}{5} e^{t}

Solution steps

x^{''} (t) + 4 x = 2 e^{t}

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{2}{5} e^{t}

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{2}{5} e^{t}

y^{^{'''}} + y = 4 e^{x}

y^{^{''}} + 4 y = 2 e^{- x}

y^{^{''}} + 2 y^{^{'}} + 10 y = 50 x, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} = \frac{10 sin ( 7 t )}{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x

What is the solution for x^{''}+4x=2e^t ?
The solution for x^{''}+4x=2e^t is x=c_{1}cos(2t)+c_{2}sin(2t)+2/5 e^t