What is the solution for y^{''}-5y^'+10y=0 ?

The solution for y^{''}-5y^'+10y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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y^{''}-5y^'+10y=0

y^{^{''}} - 5 y^{^{'}} + 10 y = 0

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

y^{''} (t) - 5 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y^{^{''}} + 36 y = 0, y^{^{'}} (0) = 2

\frac{d ^{3} y}{d x ^{3}} + y = 0

y^{^{''}} - y = 0, y (0) = 2, y^{^{'}} (0) = 0, y^{^{''}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

\frac{15}{16} u^{^{''}} + 128 H (t) = 0

What is the solution for y^{''}-5y^'+10y=0 ?
The solution for y^{''}-5y^'+10y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))