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

The solution for y^{''}-3y^'+2y=e^xsin(x) is y=c_{1}e^{2x}+c_{2}e^x+1/2 e^xcos(x)-1/2 e^xsin(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{2} e^{x} cos (x) - \frac{1}{2} e^{x} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{2} e^{x} cos (x) - \frac{1}{2} e^{x} sin (x)

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{2} e^{x} cos (x) - \frac{1}{2} e^{x} sin (x)

What is the solution for y^{''}-3y^'+2y=e^xsin(x) ?
The solution for y^{''}-3y^'+2y=e^xsin(x) is y=c_{1}e^{2x}+c_{2}e^x+1/2 e^xcos(x)-1/2 e^xsin(x)