What is the solution for y^{''}+6y^'+58y=0 ?

The solution for y^{''}+6y^'+58y=0 is y=e^{-3t}(c_{1}cos(7t)+c_{2}sin(7t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'+58y=0

y^{^{''}} + 6 y^{^{'}} + 58 y = 0

y = e^{- 3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 58 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{- 3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

\frac{d}{d x} (\frac{- x ^{2}}{8})

\int \frac{1}{( c ^{2} + x ^{2} )} d x

\frac{d}{d t} (2 e^{t})

\int \frac{11 x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

\frac{\partial}{\partial y} (- e^{- \frac{y}{x}})

What is the solution for y^{''}+6y^'+58y=0 ?
The solution for y^{''}+6y^'+58y=0 is y=e^{-3t}(c_{1}cos(7t)+c_{2}sin(7t))