What is the solution for 4y^{''}-4y^'+13y=0 ?

The solution for 4y^{''}-4y^'+13y=0 is y=e^{t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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4y^{''}-4y^'+13y=0

4 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

\frac{d y}{d x}, y = e^{4 x^{2} + 1}

\frac{\partial}{\partial y} (yz)

\frac{d}{d x} (x - 2 y)

\int \frac{x}{x + 1} d x

lo g_{10} (3 x)

What is the solution for 4y^{''}-4y^'+13y=0 ?
The solution for 4y^{''}-4y^'+13y=0 is y=e^{t/2}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))