What is the solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 is y=(1+1/26)e^{t/2}-1/26 e^{-(25t)/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 12 y^{^{'}} - \frac{25}{4} y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

Solution steps

y^{''} (t) + 12 y^{'} (t) - \frac{25}{4} y = 0

Solve linear ODE:

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

What is the solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 is y=(1+1/26)e^{t/2}-1/26 e^{-(25t)/2}