What is the solution for 4y^{''}-4y^'+50y=0 ?

The solution for 4y^{''}-4y^'+50y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))

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4y^{''}-4y^'+50y=0

4 y^{^{''}} - 4 y^{^{'}} + 50 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = 8^{x}

\int \frac{1 - 2 x - x ^{2}}{x ^{3} + x ^{2} - 3 x + 1} d x

x \to 3 lim (\frac{- x + 6}{- 2 - 1})

\frac{d}{d x} (k x e^{x})

x \to 2 lim (x^{2} + 5 x)

What is the solution for 4y^{''}-4y^'+50y=0 ?
The solution for 4y^{''}-4y^'+50y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))