What is the solution for y^{''}-8y^'+16y=3e^{-x} ?

The solution for y^{''}-8y^'+16y=3e^{-x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+3/25 e^{-x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+16y=3e^{-x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 3 e^{- x}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3}{25} e^{- x}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 3 e^{- x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3}{25} e^{- x}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3}{25} e^{- x}

y^{^{''}} + 4 y^{^{'}} + 20 y = e^{- 2 t} sin (4 t)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = x e^{x}

x^{^{''}} + 200 x^{^{'}} + 100000 x = 100 sin (100 t)

2 y^{^{''}} + 6 y^{^{'}} + 10 y = 4 cos (t)

y^{^{''}} - 14 y^{^{'}} + 49 y = 2 x e^{7 x}

What is the solution for y^{''}-8y^'+16y=3e^{-x} ?
The solution for y^{''}-8y^'+16y=3e^{-x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+3/25 e^{-x}