What is the solution for y^{''}-2y^'+5y=0,y(0)=-1,y^'(0)=3 ?

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

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y^{''}-2y^'+5y=0,y(0)=-1,y^'(0)=3

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\frac{d ^{4} y}{d t ^{4}} - y = 0

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''''}} - 8 y^{^{'}} = 0

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

y^{^{''}} + 36 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

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