What is the solution for y^{''}+36y=0,y(0)=4,y^'(0)=-5 ?

The solution for y^{''}+36y=0,y(0)=4,y^'(0)=-5 is y=4cos(6t)-5/6 sin(6t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+36y=0,y(0)=4,y^'(0)=-5

y^{^{''}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y = 4 cos (6 t) - \frac{5}{6} sin (6 t)

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

y = 4 cos (6 t) - \frac{5}{6} sin (6 t)

y = 4 cos (6 t) - \frac{5}{6} sin (6 t)

y^{^{''}} - y^{^{'}} - 72 y = 0, y (1) = e^{- 8 x}

(D^{3} - 4 D^{2} + 5 D) y = 0

y^{^{''}} + 10 y^{^{'}} - 24 y = 0

4 y^{^{''''}} - 5 y^{^{''}} - 9 y = 0

y^{^{'''}} - 3 y^{^{''}} - 34 y^{^{'}} - 48 y = 0

What is the solution for y^{''}+36y=0,y(0)=4,y^'(0)=-5 ?
The solution for y^{''}+36y=0,y(0)=4,y^'(0)=-5 is y=4cos(6t)-5/6 sin(6t)