What is the solution for y^{''}+64y=0,y(0)=1,y^'(0)=5 ?

The solution for y^{''}+64y=0,y(0)=1,y^'(0)=5 is y=cos(8t)+5/8 sin(8t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+64y=0,y(0)=1,y^'(0)=5

y^{^{''}} + 64 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = cos (8 t) + \frac{5}{8} sin (8 t)

Solution steps

y^{''} (t) + 64 y = 0

Solve linear ODE:

y = cos (8 t) + \frac{5}{8} sin (8 t)

y = cos (8 t) + \frac{5}{8} sin (8 t)

\frac{d}{d x} (- x^{3} e^{x})

\frac{\partial}{\partial x} (x e^{(x^{2} - y)})

\int (x + 2) (3 x^{2} + 12 x) d x

\frac{\partial}{\partial y} (x y^{2} + y e^{x^{2}} + 5)

y = ln (x + 5)

What is the solution for y^{''}+64y=0,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+64y=0,y(0)=1,y^'(0)=5 is y=cos(8t)+5/8 sin(8t)