What is the solution for y^{''}-4y^'+3y=2e^{(3t)} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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