What is the solution for y^{''}-9y^'+18y=e^{6t} ?

The solution for y^{''}-9y^'+18y=e^{6t} is y=c_{1}e^{6t}+c_{2}e^{3t}+(e^{6t}t)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-9y^'+18y=e^{6t}

y^{^{''}} - 9 y^{^{'}} + 18 y = e^{6 t}

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{e ^{6 t} t}{3}

Solution steps

y^{''} (t) - 9 y^{'} (t) + 18 y = e^{6 t}

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{e ^{6 t} t}{3}

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{e ^{6 t} t}{3}

y^{^{''}} + y = 2 \cdot cos (x) \cdot cos (2 x)

y^{^{''}} + y^{^{'}} - 6 y = 2 e^{- 3 x} + e^{2 x}

y^{^{''}} + 4 y = - cos (3 t) - 2 sin (3 t)

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} = 40 x - 7 - 6 e^{2 x}

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{9 x}) + 9 (cos (x))

What is the solution for y^{''}-9y^'+18y=e^{6t} ?
The solution for y^{''}-9y^'+18y=e^{6t} is y=c_{1}e^{6t}+c_{2}e^{3t}+(e^{6t}t)/3