What is the solution for y^{''}+9y=0,y(0)=2,y^'(0)=4 ?

The solution for y^{''}+9y=0,y(0)=2,y^'(0)=4 is y=2cos(3t)+4/3 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+9y=0,y(0)=2,y^'(0)=4

y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y = 2 cos (3 t) + \frac{4}{3} sin (3 t)

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

y = 2 cos (3 t) + \frac{4}{3} sin (3 t)

y = 2 cos (3 t) + \frac{4}{3} sin (3 t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = 0

u^{^{''}} + 144 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 48

y^{^{'''}} - 7 y^{^{''}} + 19 y^{^{'}} - 13 y = 0

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

y^{^{''}} + 2 y^{^{'}} + 63 y = 0

What is the solution for y^{''}+9y=0,y(0)=2,y^'(0)=4 ?
The solution for y^{''}+9y=0,y(0)=2,y^'(0)=4 is y=2cos(3t)+4/3 sin(3t)