What is the solution for 25y^{''}+80y^'+64y=0,y^'(0)=5 ?

The solution for 25y^{''}+80y^'+64y=0,y^'(0)=5 is y=-(e^{-(8t)/5}(25-5c_{2}))/(8)+c_{2}e^{-(8t)/5}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25y^{''}+80y^'+64y=0,y^'(0)=5

25 y^{^{''}} + 80 y^{^{'}} + 64 y = 0, y^{^{'}} (0) = 5

y = - \frac{e ^{- \frac{8 t}{5}} ( 25 - 5 c _{2} )}{8} + c_{2} e^{- \frac{8 t}{5}} t

Solution steps

25 y^{''} (t) + 80 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = - \frac{e ^{- \frac{8 t}{5}} ( 25 - 5 c _{2} )}{8} + c_{2} e^{- \frac{8 t}{5}} t

y = - \frac{e ^{- \frac{8 t}{5}} ( 25 - 5 c _{2} )}{8} + c_{2} e^{- \frac{8 t}{5}} t

y^{^{'''}} + 9 y^{^{''}} + 22 y^{^{'}} + 14 y = 0

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

y^{^{''}} + 36 y^{^{'}} = 0, y (0) = 1, y (0) = 0

256 y^{^{''''}} = 625 y

4 (y^{^{''}}) + 4 (y^{^{'}}) + y = 0

What is the solution for 25y^{''}+80y^'+64y=0,y^'(0)=5 ?
The solution for 25y^{''}+80y^'+64y=0,y^'(0)=5 is y=-(e^{-(8t)/5}(25-5c_{2}))/(8)+c_{2}e^{-(8t)/5}t