What is the solution for y^{'''}+2y^{''}+4y^'+8y=t^2 ?

The solution for y^{'''}+2y^{''}+4y^'+8y=t^2 is y=c_{1}e^{-2t}+c_{2}cos(2t)+c_{3}sin(2t)+(t^2)/8-t/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}+2y^{''}+4y^'+8y=t^2

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 y = t^{2}

y = c_{1} e^{- 2 t} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{2}}{8} - \frac{t}{8}

Solution steps

y^{'''} (t) + 2 y^{''} (t) + 4 y^{'} (t) + 8 y = t^{2}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{2}}{8} - \frac{t}{8}

y = c_{1} e^{- 2 t} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{2}}{8} - \frac{t}{8}

y^{^{''}} + 7 y^{^{'}} + 6 y = 30

y^{^{''}} - 4 y^{^{'}} + 13 y = sin (3 x)

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

y^{^{''}} + y^{^{'}} - 12 y = - sin (2 x) + 3 cos (2 x)

y^{^{''}} + 2 y^{^{'}} + y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 3

What is the solution for y^{'''}+2y^{''}+4y^'+8y=t^2 ?
The solution for y^{'''}+2y^{''}+4y^'+8y=t^2 is y=c_{1}e^{-2t}+c_{2}cos(2t)+c_{3}sin(2t)+(t^2)/8-t/8