What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1 ?

The solution for 4y^{''}+20y^'+25y=0,y(0)=1 is y=e^{-(5t)/2}+c_{2}e^{-(5t)/2}t

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4y^{''}+20y^'+25y=0,y(0)=1

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1

y = e^{- \frac{5 t}{2}} + c_{2} e^{- \frac{5 t}{2}} t

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} + c_{2} e^{- \frac{5 t}{2}} t

y = e^{- \frac{5 t}{2}} + c_{2} e^{- \frac{5 t}{2}} t

y^{^{''''}} - 3 y^{^{''}} - 4 y = 0

y^{^{''}} + 0.5 y^{^{'}} + y = 0

y^{^{''}} + 64 π^{2} y = 0

y^{^{''}} + 16 y^{^{'}} + 73 y = 0, y^{^{'}} (0) = 8

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 4

What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1 ?
The solution for 4y^{''}+20y^'+25y=0,y(0)=1 is y=e^{-(5t)/2}+c_{2}e^{-(5t)/2}t