What is the solution for (d^2y)/(dt^2)+16y=0,y(0)=10,y^'(0)=0 ?

The solution for (d^2y)/(dt^2)+16y=0,y(0)=10,y^'(0)=0 is y=10cos(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(d^2y)/(dt^2)+16y=0,y(0)=10,y^'(0)=0

\frac{d ^{2} y}{d t ^{2}} + 16 y = 0, y (0) = 10, y^{^{'}} (0) = 0

y = 10 cos (4 t)

Solution steps

y^{''} (t) + 16 y = 0

Solve linear ODE:

y = 10 cos (4 t)

y = 10 cos (4 t)

(D^{3} - 1) y = 0

28 y^{^{''}} + 17 y^{^{'}} - 3 y = 0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = \frac{1}{2}, x^{^{'}} (0) = 0

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{\frac{3 x}{2}}

81 y^{^{''}} - 90 y^{^{'}} + 106 y = 0

What is the solution for (d^2y)/(dt^2)+16y=0,y(0)=10,y^'(0)=0 ?
The solution for (d^2y)/(dt^2)+16y=0,y(0)=10,y^'(0)=0 is y=10cos(4t)