What is the solution for 8y^{''}-8y^'+4y=0 ?

The solution for 8y^{''}-8y^'+4y=0 is y=e^{t/2}(c_{1}cos(t/2)+c_{2}sin(t/2))

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8y^{''}-8y^'+4y=0

8 y^{^{''}} - 8 y^{^{'}} + 4 y = 0

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

Solution steps

8 y^{''} (t) - 8 y^{'} (t) + 4 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}))

\frac{d y}{d x} = 9 - \frac{3 y}{250}

Y (x) = e^{4 x + 3}

\frac{d}{d y} (y^{2} - 1)

\int \frac{1}{x ^{2} x ^{2} - 13} d x

\frac{d y}{d x} = 14 x e^{6} x (36 + y^{2})

What is the solution for 8y^{''}-8y^'+4y=0 ?
The solution for 8y^{''}-8y^'+4y=0 is y=e^{t/2}(c_{1}cos(t/2)+c_{2}sin(t/2))