What is the solution for y^{''}-y-2y=1+t ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y-2y=1+t

y^{^{''}} - y - 2 y = 1 + t

y = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{3}

Solution steps

y^{''} (t) - y - 2 y = 1 + t

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{3}

y = c_{1} e^{3 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{3}

y^{^{''}} + 5 y = g (t), y (0) = - 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 sin (2 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 e^{- 2 x}

y^{^{''}} - 4 y = 10 e^{- 2 x}

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