What is the solution for y^{''}-4y^'+13y=0,y(0)=8,y^'(0)=-2 ?

The solution for y^{''}-4y^'+13y=0,y(0)=8,y^'(0)=-2 is y=e^{2t}(8cos(3t)-6sin(3t))

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y^{''}-4y^'+13y=0,y(0)=8,y^'(0)=-2

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 8, y^{^{'}} (0) = - 2

y = e^{2 t} (8 cos (3 t) - 6 sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (8 cos (3 t) - 6 sin (3 t))

y = e^{2 t} (8 cos (3 t) - 6 sin (3 t))

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

\frac{d ^{2} f}{d x ^{2}} + f (x) = 0

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = - \frac{1}{6}

y^{^{''}} - 2 \cdot y^{^{'}} + y = 0

25 y^{^{''}} + 40 y^{^{'}} + 16 y = 0

What is the solution for y^{''}-4y^'+13y=0,y(0)=8,y^'(0)=-2 ?
The solution for y^{''}-4y^'+13y=0,y(0)=8,y^'(0)=-2 is y=e^{2t}(8cos(3t)-6sin(3t))