What is the solution for 4y^{''}+16y^'+25y=0 ?

The solution for 4y^{''}+16y^'+25y=0 is y=e^{-2t}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}+16y^'+25y=0

4 y^{^{''}} + 16 y^{^{'}} + 25 y = 0

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

4 y^{''} (t) + 16 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y^{^{''}} - 14 y^{^{'}} + 65 y = 0

y^{^{''''}} - 81 y^{^{''}} = 0

3.5 y^{^{''}} - 5.6 y^{^{'}} + 2.2 y = 0

y^{^{'''}} - 2 y^{^{'}} + 4 y = 0

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 2 y = 0

What is the solution for 4y^{''}+16y^'+25y=0 ?
The solution for 4y^{''}+16y^'+25y=0 is y=e^{-2t}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))