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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + y^{^{'}} + y = 5 x + 3

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 e^{- 2 t}

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 3 + 4 e^{- t}

y^{^{''}} + 9 y = 18 t, y (\frac{π}{2}) = 0, y (0) = 0

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