What is the solution for y^{''}-2y^'+8y=320t^2 ?

The solution for y^{''}-2y^'+8y=320t^2 is y=e^t(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))+40t^2+20t-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+8y=320t^2

y^{^{''}} - 2 y^{^{'}} + 8 y = 320 t^{2}

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + 40 t^{2} + 20 t - 5

Solution steps

y^{''} (t) - 2 y^{'} (t) + 8 y = 320 t^{2}

Solve linear ODE:

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + 40 t^{2} + 20 t - 5

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + 40 t^{2} + 20 t - 5

\frac{\partial}{\partial z} (- 2 x z)

\frac{1}{3 x}

x \to 4 lim (\frac{x + 4}{x - 4})

\int \frac{t}{e ^{3 t}} d t

\int 3 e^{- 4 x} sin (e^{- 4 x}) d x

What is the solution for y^{''}-2y^'+8y=320t^2 ?
The solution for y^{''}-2y^'+8y=320t^2 is y=e^t(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))+40t^2+20t-5