What is the solution for y^{''}+49y=0,y(0)=6,y^'(0)=-6 ?

The solution for y^{''}+49y=0,y(0)=6,y^'(0)=-6 is y=6cos(7t)-6/7 sin(7t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+49y=0,y(0)=6,y^'(0)=-6

y^{^{''}} + 49 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

y = 6 cos (7 t) - \frac{6}{7} sin (7 t)

Solution steps

y^{''} (t) + 49 y = 0

Solve linear ODE:

y = 6 cos (7 t) - \frac{6}{7} sin (7 t)

y = 6 cos (7 t) - \frac{6}{7} sin (7 t)

3 y^{^{'''}} + 11 y^{^{''}} + 24 y^{^{'}} + 12 y = 0

(D^{2} + 2 D + 17) y = 0

3 y^{^{'''}} - 5 y^{^{''}} - 16 y^{^{'}} + 12 y = 0

y^{^{''}} + y^{^{'}} + 4.25 y = 0, y (0) = 5

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

What is the solution for y^{''}+49y=0,y(0)=6,y^'(0)=-6 ?
The solution for y^{''}+49y=0,y(0)=6,y^'(0)=-6 is y=6cos(7t)-6/7 sin(7t)