What is the solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=-5 ?

The solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=-5 is y=e^{2t}(2cos(t)-9sin(t))

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y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=-5

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y = e^{2 t} (2 cos (t) - 9 sin (t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{2 t} (2 cos (t) - 9 sin (t))

y = e^{2 t} (2 cos (t) - 9 sin (t))

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

(D^{3} - 6 D^{2} + 11 D - 6) y = 0

20 y^{^{''}} + 240 y^{^{'}} + 800 y = 0

a y^{^{''}} + b y^{^{'}} - cy = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 3, y (0) = 5

What is the solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=-5 ?
The solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=-5 is y=e^{2t}(2cos(t)-9sin(t))