What is the solution for y^{''}-6y^'+10y=0,y(0)=-1,y^'(0)=3 ?

The solution for y^{''}-6y^'+10y=0,y(0)=-1,y^'(0)=3 is y=e^{3t}(-cos(t)+6sin(t))

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y^{''}-6y^'+10y=0,y(0)=-1,y^'(0)=3

y^{^{''}} - 6 y^{^{'}} + 10 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

y = e^{3 t} (- cos (t) + 6 sin (t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{3 t} (- cos (t) + 6 sin (t))

y = e^{3 t} (- cos (t) + 6 sin (t))

20 y^{^{''}} - 100 y^{^{'}} + 133 y = 0

y^{^{'''}} + 9 y^{^{'}} = 0, y^{^{''}} (0) = 27

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 1, x^{^{'}} (0) = 2

y^{^{''}} - 10 y^{^{'}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} - 20 y = 0

What is the solution for y^{''}-6y^'+10y=0,y(0)=-1,y^'(0)=3 ?
The solution for y^{''}-6y^'+10y=0,y(0)=-1,y^'(0)=3 is y=e^{3t}(-cos(t)+6sin(t))