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

The solution for y^{''}-y^'-2y=8e-t is y=c_{1}e^{2t}+c_{2}e^{-t}+t/2-(16e+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=8e-t

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

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{t}{2} - \frac{16 e + 1}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{t}{2} - \frac{16 e + 1}{4}

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{t}{2} - \frac{16 e + 1}{4}

f (x) = x^{\frac{1}{x}}

\frac{\partial}{\partial y} (y + y cos (x y))

\frac{d r}{d x} = 4 π r^{2}

\frac{1}{2 x + 25}

\int \frac{19}{x 1 + 2 x} d x

What is the solution for y^{''}-y^'-2y=8e-t ?
The solution for y^{''}-y^'-2y=8e-t is y=c_{1}e^{2t}+c_{2}e^{-t}+t/2-(16e+1)/4