What is the solution for y^{''}-3y^'+2y=4e^{3x} ?

The solution for y^{''}-3y^'+2y=4e^{3x} is y=c_{1}e^{2x}+c_{2}e^x+2e^{3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 4 e^{3 x}

Solve linear ODE:

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

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

What is the solution for y^{''}-3y^'+2y=4e^{3x} ?
The solution for y^{''}-3y^'+2y=4e^{3x} is y=c_{1}e^{2x}+c_{2}e^x+2e^{3x}