What is the solution for y^{''}+16y=0,y^'(0)=3,y(0)=5 ?

The solution for y^{''}+16y=0,y^'(0)=3,y(0)=5 is y=5cos(4t)+3/4 sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=0,y^'(0)=3,y(0)=5

y^{^{''}} + 16 y = 0, y^{^{'}} (0) = 3, y (0) = 5

y = 5 cos (4 t) + \frac{3}{4} sin (4 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (4 t) + \frac{3}{4} sin (4 t)

y = 5 cos (4 t) + \frac{3}{4} sin (4 t)

5 y^{^{''}} + 25 y = 0

y^{^{''}} - 15 y^{^{'}} + 54 y = 0, y (0) = 9, y^{^{'}} (0) = 6

- x^{^{''}} + 2 x^{^{'}} + x = 0

\frac{d ^{4} y}{d x ^{4}} + 4 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 4

What is the solution for y^{''}+16y=0,y^'(0)=3,y(0)=5 ?
The solution for y^{''}+16y=0,y^'(0)=3,y(0)=5 is y=5cos(4t)+3/4 sin(4t)