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

The solution for y^{'''}+2y^'=cos(x) is y=c_{1}+c_{2}cos(sqrt(2)x)+c_{3}sin(sqrt(2)x)+sin(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + sin (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + sin (x)

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{2 x}) + 5 cos (x)

(D^{4} - 1) y = e^{- x}

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

y^{^{''}} + 4 y = 3 cos (2.1 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x}, y (\frac{π}{2}) = 0, y (0) = 0

What is the solution for y^{'''}+2y^'=cos(x) ?
The solution for y^{'''}+2y^'=cos(x) is y=c_{1}+c_{2}cos(sqrt(2)x)+c_{3}sin(sqrt(2)x)+sin(x)