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

The solution for y^{''}-4y^'+4y=2e^{2t}+4t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+e^{2t}t^2+t^2+2t+3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 8 y = e^{2 x}

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

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (5 t)

y^{^{''}} + y^{^{'}} - 2 y = 2 x, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 16 y = 1 - x + e^{4 x}

What is the solution for y^{''}-4y^'+4y=2e^{2t}+4t^2 ?
The solution for y^{''}-4y^'+4y=2e^{2t}+4t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+e^{2t}t^2+t^2+2t+3/2