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

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

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4y^{''}+20y^'+25y=0

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0

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

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

y = e^{x}, y = 0, x = 0, x = ln (3)

\int e^{x} 8 - e^{x} d x

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a) + cos^{2} (a))

\frac{d}{d x} (x (x + 1))

\frac{d}{d x} (4 x^{3} - \frac{2}{x ^{3}})

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