What is the solution for y^{''}-2y^'+53/4 y=0 ?

The solution for y^{''}-2y^'+53/4 y=0 is y=e^t(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))

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y^{''}-2y^'+53/4 y=0

y^{^{''}} - 2 y^{^{'}} + \frac{53}{4} y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + \frac{53}{4} y = 0

Solve linear ODE:

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

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

x \to 0 - lim (\frac{cos ( x )}{x})

\frac{d y}{d x} = \frac{( 2 x + sec ^{2} ( x ) )}{2 y}

x \to 2 + lim (\frac{x - 13}{- x + 2})

(t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\int (3^{2 x} + 3^{4 x})^{2} d x

What is the solution for y^{''}-2y^'+53/4 y=0 ?
The solution for y^{''}-2y^'+53/4 y=0 is y=e^t(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))