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

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

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

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

h (t) = e^{t} (cos (3 t) + 3 sin (3 t))

Solution steps

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

Solve linear ODE:

h (t) = e^{t} (cos (3 t) + 3 sin (3 t))

h (t) = e^{t} (cos (3 t) + 3 sin (3 t))

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

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

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} = 0

64 y^{^{''}} + 144 y^{^{'}} + 81 y = 0

y^{^{'''}} + 3 y^{^{''}} - 22 y^{^{'}} - 24 y = 0

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