What is the solution for 10y^{''}-10y^'+25y=0 ?

The solution for 10y^{''}-10y^'+25y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

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10y^{''}-10y^'+25y=0

10 y^{^{''}} - 10 y^{^{'}} + 25 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

10 y^{''} (t) - 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

f (x) = (\frac{2 + 3 x}{3 x}) (9 - 4 x)

\frac{d}{d x} (5 sin (x) cos (x))

\int (x + 1) (9 x - 8) d x

\int \frac{x ^{7} - 4 x}{x ^{5}} d x

sin (10 x)

What is the solution for 10y^{''}-10y^'+25y=0 ?
The solution for 10y^{''}-10y^'+25y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))