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

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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

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