What is the solution for 4y^{''}-4y^'+2y=0 ?

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

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4y^{''}-4y^'+2y=0

4 y^{^{''}} - 4 y^{^{'}} + 2 y = 0

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 2 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{x}{x - 1}

\int \frac{x}{e ^{7 x}} d x

x \to 1 lim (\frac{2}{5} x + 7)

\int ln (3)

\frac{\partial}{\partial x} (\frac{6}{x ^{2}})

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