What is the solution for y^{''}+4y^'-32y=0,y(0)=-3 ?

The solution for y^{''}+4y^'-32y=0,y(0)=-3 is y=(-3-c_{2})e^{4t}+c_{2}e^{-8t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'-32y=0,y(0)=-3

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = - 3

y = (- 3 - c_{2}) e^{4 t} + c_{2} e^{- 8 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 32 y = 0

Solve linear ODE:

y = (- 3 - c_{2}) e^{4 t} + c_{2} e^{- 8 t}

y = (- 3 - c_{2}) e^{4 t} + c_{2} e^{- 8 t}

x^{^{'''}} + x^{^{''}} - 72 x^{^{'}} = 0

y^{^{''''}} = 256 y

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

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

2 y^{^{''}} - 3 y^{^{'}} + 8 y = 0

What is the solution for y^{''}+4y^'-32y=0,y(0)=-3 ?
The solution for y^{''}+4y^'-32y=0,y(0)=-3 is y=(-3-c_{2})e^{4t}+c_{2}e^{-8t}