What is the solution for y^{''}-2y^'+10y=0,y(0)=1,y^'(0)=5 ?

The solution for y^{''}-2y^'+10y=0,y(0)=1,y^'(0)=5 is y=e^t(cos(3t)+4/3 sin(3t))

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y^{''}-2y^'+10y=0,y(0)=1,y^'(0)=5

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = e^{t} (cos (3 t) + \frac{4}{3} sin (3 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{t} (cos (3 t) + \frac{4}{3} sin (3 t))

y = e^{t} (cos (3 t) + \frac{4}{3} sin (3 t))

y^{^{''}} + 4 y = 0, y (\frac{π}{6}) = 5, y (0) = - 2

y^{^{''''}} - 14 y^{^{'''}} + 49 y^{^{''}} = 0

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

y^{^{''}} - 12 y^{^{'}} + 45 y = 0

x^{^{''}} + 3 x^{^{'}} + 4 x = 0

What is the solution for y^{''}-2y^'+10y=0,y(0)=1,y^'(0)=5 ?
The solution for y^{''}-2y^'+10y=0,y(0)=1,y^'(0)=5 is y=e^t(cos(3t)+4/3 sin(3t))