What is the solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 ?

The solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 is y=(1+13/5)e^{t/2}-13/5 e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 7

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y^{^{'}} (0) = 1

5 y^{^{''}} + 9 y^{^{'}} - 14 y = 0

4 y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = - 3

y^{^{''}} - 2 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = - 6

What is the solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 ?
The solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 is y=(1+13/5)e^{t/2}-13/5 e^{-2t}