What is the solution for 2y^{''}+7y^'+3y=0,y^'(0)=0,y(0)=3 ?

The solution for 2y^{''}+7y^'+3y=0,y^'(0)=0,y(0)=3 is y= 18/5 e^{-t/2}-3/5 e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+7y^'+3y=0,y^'(0)=0,y(0)=3

2 y^{^{''}} + 7 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 0, y (0) = 3

y = \frac{18}{5} e^{- \frac{t}{2}} - \frac{3}{5} e^{- 3 t}

Solution steps

2 y^{''} (t) + 7 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = \frac{18}{5} e^{- \frac{t}{2}} - \frac{3}{5} e^{- 3 t}

y = \frac{18}{5} e^{- \frac{t}{2}} - \frac{3}{5} e^{- 3 t}

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos (2 x)

y^{^{''}} = \frac{( a - 2 )}{a} y

16 y^{^{''}} - 32 y^{^{'}} + 65 y = 0

4 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 9

What is the solution for 2y^{''}+7y^'+3y=0,y^'(0)=0,y(0)=3 ?
The solution for 2y^{''}+7y^'+3y=0,y^'(0)=0,y(0)=3 is y= 18/5 e^{-t/2}-3/5 e^{-3t}