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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 16 y^{^{'}} = e^{- π x}

x^{^{''}} + 4 x = 4 sec^{2} (2 t)

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x} cos (2 x)

y^{^{''}} + y = cos (- t + \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 1

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