What is the solution for y^{''}+32y^'+224y=0 ?

The solution for y^{''}+32y^'+224y=0 is y=c_{1}e^{4(sqrt(2)-4)t}+c_{2}e^{-4(4+sqrt(2))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+32y^'+224y=0

y^{^{''}} + 32 y^{^{'}} + 224 y = 0

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

Solution steps

y^{''} (t) + 32 y^{'} (t) + 224 y = 0

Solve linear ODE:

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

x^{^{''}} = x \cdot 5.3

y^{^{''}} = 2 y^{^{'}} - y

5 y^{^{''}} - 15 y = 0

y^{^{''}} + 8 y^{^{'}} + 210 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

(\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y)^{3} = 0

What is the solution for y^{''}+32y^'+224y=0 ?
The solution for y^{''}+32y^'+224y=0 is y=c_{1}e^{4(sqrt(2)-4)t}+c_{2}e^{-4(4+sqrt(2))t}