What is the solution for y^{''}-6y^'+10y=0,y(0)=2,y^'(0)=3 ?

The solution for y^{''}-6y^'+10y=0,y(0)=2,y^'(0)=3 is y=e^{3t}(2cos(t)-3sin(t))

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y^{''}-6y^'+10y=0,y(0)=2,y^'(0)=3

y^{^{''}} - 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = e^{3 t} (2 cos (t) - 3 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{3 t} (2 cos (t) - 3 sin (t))

y = e^{3 t} (2 cos (t) - 3 sin (t))

7 x^{^{''}} + 3 x^{^{''}} + 7 x = 0, x (0) = - 4, x^{^{'}} (0) = 0

y^{^{'''}} + y^{^{''}} - 56 y^{^{'}} = 0

18 y^{^{''}} + (45 - 27) y^{^{'}} - 57 y = 0

y^{^{''}} - 10 y + 24 y = 0

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

What is the solution for y^{''}-6y^'+10y=0,y(0)=2,y^'(0)=3 ?
The solution for y^{''}-6y^'+10y=0,y(0)=2,y^'(0)=3 is y=e^{3t}(2cos(t)-3sin(t))