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

The solution for y^{''}-2y^'+2y=sin(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))+1/5 sin(x)+2/5 cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y^{^{''}} = 1 - y, y (0) = 0, y (1) = 0

(4 d^{2} + 12 d + 9) y = 144 e^{- 3} x

y^{^{''}} + y^{^{'}} = x^{2} + 2 x + 4

y^{^{''}} + 2 y = x + 1

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 = 0

What is the solution for y^{''}-2y^'+2y=sin(x) ?
The solution for y^{''}-2y^'+2y=sin(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))+1/5 sin(x)+2/5 cos(x)