What is the solution for y^{''}-4y^'+13y=0,y(0)=1,y^'(0)=4 ?

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

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y^{''}-4y^'+13y=0,y(0)=1,y^'(0)=4

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

Solution steps

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

Solve linear ODE:

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

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

\int x^{2} + 1 x d x

x \to 0 lim (sin (x) \cdot ln (4 x))

\frac{\partial}{\partial x} (2 x + y - 5)

\frac{d}{d x} (arccos (\frac{y}{x}))

y^{^{'}} = y \cdot tan (x) + sin (x)

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