What is the solution for y^{''}+6*0+16y=0 ?

The solution for y^{''}+6*0+16y=0 is y=c_{1}cos(4t)+c_{2}sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6*0+16y=0

y^{^{''}} + 6 \cdot 0 + 16 y = 0

y = c_{1} cos (4 t) + c_{2} sin (4 t)

Solution steps

y^{''} (t) + 6 \cdot 0 + 16 y = 0

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t)

- 4 y^{^{''}} + y^{^{'}} = 0

y^{^{''}} - \frac{1}{r ^{2}} y = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (1) = e^{2 x}

y^{^{''}} + 15 y^{^{'}} + 56 y = 0, y (0) = 11, y^{^{'}} (0) = - 85

4 y^{^{''}} = - 25 y

What is the solution for y^{''}+6*0+16y=0 ?
The solution for y^{''}+6*0+16y=0 is y=c_{1}cos(4t)+c_{2}sin(4t)