What is the solution for y^{''}=by ?

The solution for y^{''}=by is y=c_{1}e^{sqrt(b)t}+c_{2}e^{-sqrt(b)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}=by

y^{^{''}} = b y

y = c_{1} e^{b t} + c_{2} e^{- b t}

Solution steps

y^{''} (t) = b y (t)

Solve linear ODE:

y = c_{1} e^{b t} + c_{2} e^{- b t}

y = c_{1} e^{b t} + c_{2} e^{- b t}

\int_{1}^{2} \frac{x ^{2} + 5}{3 x - x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{4 + x})

\int_{0}^{\frac{3}{4}} 60 x d x

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

x \to \infty lim (x^{2} - 4)

What is the solution for y^{''}=by ?
The solution for y^{''}=by is y=c_{1}e^{sqrt(b)t}+c_{2}e^{-sqrt(b)t}