What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=6 ?

The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=6 is y=e^{2t}(cos(t)+4sin(t))

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y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=6

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 6

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

4 y^{^{''}} - 24 y^{^{'}} + 37 y = 0, y (0) = - 9

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

8 y^{^{''}} + 16 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

(d^{2} + d + 1) y = 0

What is the solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=6 ?
The solution for y^{''}-4y^'+5y=0,y(0)=1,y^'(0)=6 is y=e^{2t}(cos(t)+4sin(t))