What is the solution for y^{''}-3y^'-18y=xe^{3x} ?

The solution for y^{''}-3y^'-18y=xe^{3x} is y=c_{1}e^{6x}+c_{2}e^{-3x}-(e^{3x}x)/(18)-1/108 e^{3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'-18y=xe^{3x}

y^{^{''}} - 3 y^{^{'}} - 18 y = x e^{3 x}

y = c_{1} e^{6 x} + c_{2} e^{- 3 x} - \frac{e ^{3 x} x}{18} - \frac{1}{108} e^{3 x}

Solution steps

y^{''} (x) - 3 y^{'} (x) - 18 y = x e^{3 x}

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 3 x} - \frac{e ^{3 x} x}{18} - \frac{1}{108} e^{3 x}

y = c_{1} e^{6 x} + c_{2} e^{- 3 x} - \frac{e ^{3 x} x}{18} - \frac{1}{108} e^{3 x}

y^{^{''}} + y = e^{- t} sin (e^{- t}) + cos (e^{- t})

y^{^{''}} + y = \frac{x}{∣ x ∣}

(D^{2} + 5 D + 6) y = x^{2} + 2 x

2 y^{^{'''}} + 9 y^{^{''}} + y^{^{'}} - 8 y = e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 4 x - 5 + 6 x e^{x}

What is the solution for y^{''}-3y^'-18y=xe^{3x} ?
The solution for y^{''}-3y^'-18y=xe^{3x} is y=c_{1}e^{6x}+c_{2}e^{-3x}-(e^{3x}x)/(18)-1/108 e^{3x}