What is the solution for 16y^{''}+40y^'+25y=0 ?

The solution for 16y^{''}+40y^'+25y=0 is y=c_{1}e^{-(5t)/4}+c_{2}e^{-(5t)/4}t

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16y^{''}+40y^'+25y=0

16 y^{^{''}} + 40 y^{^{'}} + 25 y = 0

y = c_{1} e^{- \frac{5 t}{4}} + c_{2} e^{- \frac{5 t}{4}} t

Solution steps

16 y^{''} (t) + 40 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{5 t}{4}} + c_{2} e^{- \frac{5 t}{4}} t

y = c_{1} e^{- \frac{5 t}{4}} + c_{2} e^{- \frac{5 t}{4}} t

\int (x + 4) e^{x^{2} + 8 x} d x

(6 x + 1) y^{2} (\frac{d y}{d x}) + 3 x^{2} + 2 y^{3} = 0

y = x^{2} cos (x)

\frac{d y}{d x} = 4 x 4 - y^{2}

\int x sin (ln (x)) d x

What is the solution for 16y^{''}+40y^'+25y=0 ?
The solution for 16y^{''}+40y^'+25y=0 is y=c_{1}e^{-(5t)/4}+c_{2}e^{-(5t)/4}t