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

The solution for y^{''}-2y^'+2y=4e^xcos(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))+2e^xxsin(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + y = sin (x) + x cos (x)

y^{^{''}} + 2 y^{^{'}} - 5 y = - 2

4 y^{^{''}} + 16 y = 1, y (0) = - 2, y^{^{'}} (0) = 0

y^{^{''}} + y = t^{3}

What is the solution for y^{''}-2y^'+2y=4e^xcos(x) ?
The solution for y^{''}-2y^'+2y=4e^xcos(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))+2e^xxsin(x)