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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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y^{^{''}} - 2 y^{^{'}} - 3 y = t

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