What is the solution for y^{''}-4y^'+4y=2e^{5t} ?

The solution for y^{''}-4y^'+4y=2e^{5t} is y=c_{1}e^{2t}+c_{2}te^{2t}+2/9 e^{5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=2e^{5t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{5 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{2}{9} e^{5 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 2 e^{5 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{2}{9} e^{5 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{2}{9} e^{5 t}

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88

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

\frac{d ^{2} y}{d x ^{2}} - 25 y = 3 x^{2} - 2 + e^{x}

\frac{d ^{2} y}{d x ^{2}} - 4 (\frac{d y}{d x}) + 8 y = x e^{x}

y^{^{'''}} + 4 y^{^{''}} - 2 y = 2 x^{2} - 3 x + 6

What is the solution for y^{''}-4y^'+4y=2e^{5t} ?
The solution for y^{''}-4y^'+4y=2e^{5t} is y=c_{1}e^{2t}+c_{2}te^{2t}+2/9 e^{5t}