What is the solution for y^{''}-y^'-2y=2e^x ?

The solution for y^{''}-y^'-2y=2e^x is y=c_{1}e^{2x}+c_{2}e^{-x}-e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- x} - e^{x}

(D^{2} - 4 D + 13) y = 24 e^{2 x} \cdot cos (3 x)

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

y^{^{''}} + 36 y = 11 sec^{2} (6 t)

y^{^{''}} + y^{^{'}} - 6 y = 50 x e^{2 x}

y^{^{''}} + 6 y^{^{'}} - 32 y = - 120 sin (4 t)

What is the solution for y^{''}-y^'-2y=2e^x ?
The solution for y^{''}-y^'-2y=2e^x is y=c_{1}e^{2x}+c_{2}e^{-x}-e^x