What is the solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=-5 ?

The solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=-5 is y=e^{2t}(-cos(3t)-sin(3t))

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y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=-5

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

y = e^{2 t} (- cos (3 t) - sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (- cos (3 t) - sin (3 t))

y = e^{2 t} (- cos (3 t) - sin (3 t))

f (x) = arccos (e^{2 x})

x^{3} - 8 x^{2} + 12 x, - x^{3} + 8 x^{2} - 12 x

y = \frac{7}{( 1 - x ^{2} ) ^{3}}

\frac{\partial}{\partial x} (\frac{1}{x ^{4}})

e^{x} (y - 1) d x + 2 (e^{x} + 4) d y = 0

What is the solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=-5 ?
The solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=-5 is y=e^{2t}(-cos(3t)-sin(3t))