What is the solution for y^{''}+144y=0,y(0)=3,y^'(0)=-48 ?

The solution for y^{''}+144y=0,y(0)=3,y^'(0)=-48 is y=3cos(12t)-4sin(12t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+144y=0,y(0)=3,y^'(0)=-48

y^{^{''}} + 144 y = 0, y (0) = 3, y^{^{'}} (0) = - 48

y = 3 cos (12 t) - 4 sin (12 t)

Solution steps

y^{''} (t) + 144 y = 0

Solve linear ODE:

y = 3 cos (12 t) - 4 sin (12 t)

y = 3 cos (12 t) - 4 sin (12 t)

y^{^{''}} + 4 y^{^{'}} - 12 y = 0, y (0) = 7, y^{^{'}} (0) = - 7

y^{^{''}} + 3 y^{^{'}} - y = 0

x^{^{''}} - 5 x^{^{'}} + 6 x = 0

y^{^{''}} - 11 y^{^{'}} + 28 y = 0

y^{^{'''}} + 2 y^{^{''}} = 0

What is the solution for y^{''}+144y=0,y(0)=3,y^'(0)=-48 ?
The solution for y^{''}+144y=0,y(0)=3,y^'(0)=-48 is y=3cos(12t)-4sin(12t)