What is the solution for y^{''}-2y^'+65y=0,y(0)=1,y^'(0)=4 ?

The solution for y^{''}-2y^'+65y=0,y(0)=1,y^'(0)=4 is y=e^t(cos(8t)+3/8 sin(8t))

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y^{''}-2y^'+65y=0,y(0)=1,y^'(0)=4

y^{^{''}} - 2 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y = e^{t} (cos (8 t) + \frac{3}{8} sin (8 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 65 y = 0

Solve linear ODE:

y = e^{t} (cos (8 t) + \frac{3}{8} sin (8 t))

y = e^{t} (cos (8 t) + \frac{3}{8} sin (8 t))

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (4) = 2, y^{^{'}} (4) = 6

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

What is the solution for y^{''}-2y^'+65y=0,y(0)=1,y^'(0)=4 ?
The solution for y^{''}-2y^'+65y=0,y(0)=1,y^'(0)=4 is y=e^t(cos(8t)+3/8 sin(8t))