What is the solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=4 ?

The solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=4 is y= 80/3 e^{-(3t)/5}-80/3 e^{-(3t)/4}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=4

20 y^{^{''}} + 27 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 4

y = \frac{80}{3} e^{- \frac{3 t}{5}} - \frac{80}{3} e^{- \frac{3 t}{4}}

Solution steps

20 y^{''} (t) + 27 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = \frac{80}{3} e^{- \frac{3 t}{5}} - \frac{80}{3} e^{- \frac{3 t}{4}}

y = \frac{80}{3} e^{- \frac{3 t}{5}} - \frac{80}{3} e^{- \frac{3 t}{4}}

What is the solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=4 ?
The solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=4 is y= 80/3 e^{-(3t)/5}-80/3 e^{-(3t)/4}