What is the solution for y^{''}-4y^'+29y=0,y(0)=4,y^'(0)=18 ?

The solution for y^{''}-4y^'+29y=0,y(0)=4,y^'(0)=18 is y=e^{2t}(4cos(5t)+2sin(5t))

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y^{''}-4y^'+29y=0,y(0)=4,y^'(0)=18

y^{^{''}} - 4 y^{^{'}} + 29 y = 0, y (0) = 4, y^{^{'}} (0) = 18

y = e^{2 t} (4 cos (5 t) + 2 sin (5 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{2 t} (4 cos (5 t) + 2 sin (5 t))

y = e^{2 t} (4 cos (5 t) + 2 sin (5 t))

y^{^{''}} - 83 y^{^{'}} + α y = 0

(D^{2} - 2 D + 4) y = 0

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

y^{^{''}} + 0 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 1

What is the solution for y^{''}-4y^'+29y=0,y(0)=4,y^'(0)=18 ?
The solution for y^{''}-4y^'+29y=0,y(0)=4,y^'(0)=18 is y=e^{2t}(4cos(5t)+2sin(5t))