What is the solution for 8y^{''}-8y^'+20y=0 ?

The solution for 8y^{''}-8y^'+20y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

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8y^{''}-8y^'+20y=0

8 y^{^{''}} - 8 y^{^{'}} + 20 y = 0

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

Solution steps

8 y^{''} (t) - 8 y^{'} (t) + 20 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}))

y^{^{''}} + 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d t ^{2}} + 16 y = 0

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 8 x = 0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 8 x = 0

x \cdot y^{^{''}} + 2 x \cdot y^{^{'}} + x \cdot y = 0

What is the solution for 8y^{''}-8y^'+20y=0 ?
The solution for 8y^{''}-8y^'+20y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))