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

The solution for x^{''}-4x^'+4x=e^{2t} is x=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-4x^'+4x=e^{2t}

x^{^{''}} - 4 x^{^{'}} + 4 x = e^{2 t}

x = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{2}}{2}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{2}}{2}

x = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{2}}{2}

y^{^{''}} + 4 y = 4 cos (2 x) + 6 cos (x) + 8 x^{2} - 4 x

y^{^{''}} - 4 y^{^{'}} + 3 y = 20 cos (x)

x^{^{''}} + x^{^{'}} + 12 x = 20 cos (3 t)

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 e^{- 3 t}

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

What is the solution for x^{''}-4x^'+4x=e^{2t} ?
The solution for x^{''}-4x^'+4x=e^{2t} is x=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^2)/2