What is the solution for y^{''}+4y=0.5 ?

The solution for y^{''}+4y=0.5 is y=c_{1}cos(2t)+c_{2}sin(2t)+0.125

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=0.5

y^{^{''}} + 4 y = 0.5

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 0.125

Solution steps

y^{''} (t) + 4 y = 0.5

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 0.125

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 0.125

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

What is the solution for y^{''}+4y=0.5 ?
The solution for y^{''}+4y=0.5 is y=c_{1}cos(2t)+c_{2}sin(2t)+0.125