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

The solution for y^{''}+4y=2e^t is y=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 >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 2 \cdot y = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( 16 e ^{- 3 x} )}{x ^{2} + 1}

y^{^{''}} - 2 y^{^{'}} - 24 y = 72 e^{2 t}

y^{^{''}} + 49 y = t^{2} cos (7 t) + t e^{(- 7 t)}

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

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