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

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

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^{2 t} + c_{2} e^{- t} - \frac{t}{2} - \frac{1}{4}

Solution steps

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

Solve linear ODE:

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

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

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

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

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

x^{^{''}} + 8 x^{^{'}} + 16 x = 4 sin (4 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y (0) = 1

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