What is the solution for y^{''}+y=-8xcos(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+y=-8xcos(x)

y^{^{''}} + y = - 8 x cos (x)

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

Solution steps

y^{''} (x) + y = - 8 x cos (x)

Solve linear ODE:

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

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

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