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

The solution for y^{''}-y^'-2y=-8t+8t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-4t^2+8t-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 8 t - 8

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 8 t - 8

y = c_{1} e^{2 t} + c_{2} e^{- t} - 4 t^{2} + 8 t - 8

\int e^{2 t} cos (5 t) d t

(x^{2} + 1) \frac{d y}{d x} = 2 x y

x \to 0 + lim (x^{2} ln (x))

x \to (1) lim (\frac{x ^{8} - 1}{x ^{5} - 1})

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

What is the solution for y^{''}-y^'-2y=-8t+8t^2 ?
The solution for y^{''}-y^'-2y=-8t+8t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-4t^2+8t-8