What is the solution for y^{''}+9y=18t,y(pi/2)=0 ?

The solution for y^{''}+9y=18t,y(pi/2)=0 is y=c_{1}cos(3t)+pisin(3t)+2t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+9y=18t,y(pi/2)=0

y^{^{''}} + 9 y = 18 t, y (\frac{π}{2}) = 0

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

Solution steps

y^{''} (t) + 9 y = 18 t

Solve linear ODE:

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

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

y^{^{''}} + 16 y = csc^{2} (4 t)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{- 3 x}

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

y^{^{''}} - y^{^{'}} - 2 y = 6 x + 5 e^{4 x}

y^{^{''}} - 4 y^{^{'}} - 12 y = e^{- 2 x} + 5 x + 2

What is the solution for y^{''}+9y=18t,y(pi/2)=0 ?
The solution for y^{''}+9y=18t,y(pi/2)=0 is y=c_{1}cos(3t)+pisin(3t)+2t