What is the solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=0 ?

The solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=0 is y=e^{-t}(3cos(4t)+3/4 sin(4t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y = e^{- t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{- t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

y = e^{- t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

x \to 2 lim (ln (4 - 2 x))

\int 1 + x^{2} x d x

\frac{d}{d x} (- 2 x - 3 x^{5})

\frac{d}{d x} (\frac{4 x + 9}{x})

\frac{( x ^{2} - 1 ) ^{3}}{x ^{2} + 4}

What is the solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=0 ?
The solution for y^{''}+2y^'+17y=0,y(0)=3,y^'(0)=0 is y=e^{-t}(3cos(4t)+3/4 sin(4t))