What is the solution for y^{''}+4y^'+5y=4+12t-5t^2 ?

The solution for y^{''}+4y^'+5y=4+12t-5t^2 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))-t^2+4t-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+5y=4+12t-5t^2

y^{^{''}} + 4 y^{^{'}} + 5 y = 4 + 12 t - 5 t^{2}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) - t^{2} + 4 t - 2

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 4 + 12 t - 5 t^{2}

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) - t^{2} + 4 t - 2

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) - t^{2} + 4 t - 2

y^{^{''}} + 7 y^{^{'}} = 196 sin (7 t) + 588 cos (7 t)

x^{^{''}} + 49 x = 96 cos (5 t)

\frac{d ^{2} y}{d x ^{2}} - y = x e^{2 x}

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x

What is the solution for y^{''}+4y^'+5y=4+12t-5t^2 ?
The solution for y^{''}+4y^'+5y=4+12t-5t^2 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))-t^2+4t-2