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

The solution for y^{''}+y=2xsin(2x) is y=c_{1}cos(x)+c_{2}sin(x)-(2xsin(2x))/3-8/9 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} cos (x) + c_{2} sin (x) - \frac{2 x sin ( 2 x )}{3} - \frac{8}{9} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{2 x sin ( 2 x )}{3} - \frac{8}{9} cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{2 x sin ( 2 x )}{3} - \frac{8}{9} cos (2 x)

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = x + 1, y (0) = 3, y^{^{'}} (0) = 0

d, d (t) = 4000 - 3000 e - 0.06 t

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

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

What is the solution for y^{''}+y=2xsin(2x) ?
The solution for y^{''}+y=2xsin(2x) is y=c_{1}cos(x)+c_{2}sin(x)-(2xsin(2x))/3-8/9 cos(2x)