What is the solution for 5x^{''}+10x^'+25x=0 ?

The solution for 5x^{''}+10x^'+25x=0 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))

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5x^{''}+10x^'+25x=0

5 x^{^{''}} + 10 x^{^{'}} + 25 x = 0

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

5 x^{''} (t) + 10 x^{'} (t) + 25 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t))

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t))

u^{^{''}} + 7 u^{^{'}} + 2 H (t) = 0

y^{^{''}} + 6 y^{^{'}} - 16 y = 7 y^{^{'}} - 10 y

y^{^{''}} - \frac{n ^{2}}{4} y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 (\frac{d y}{d x}) + 4 y = 0

4 x^{^{''}} + 3 x^{^{'}} + 5 x = 0

What is the solution for 5x^{''}+10x^'+25x=0 ?
The solution for 5x^{''}+10x^'+25x=0 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))