What is the solution for y^{''}-6y^'+25y=0,y(0)=-3,y^'(0)=-1 ?

The solution for y^{''}-6y^'+25y=0,y(0)=-3,y^'(0)=-1 is y=e^{3t}(-3cos(4t)+2sin(4t))

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y^{''}-6y^'+25y=0,y(0)=-3,y^'(0)=-1

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = - 1

y = e^{3 t} (- 3 cos (4 t) + 2 sin (4 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{3 t} (- 3 cos (4 t) + 2 sin (4 t))

y = e^{3 t} (- 3 cos (4 t) + 2 sin (4 t))

y^{^{''}} + \frac{100}{1} y^{^{'}} = 0

y^{^{''}} + 3 y^{^{'}} + (1 + a) y = 0

y^{^{''}} = 2 y + y^{^{'}}

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 23, y (1) = 0

What is the solution for y^{''}-6y^'+25y=0,y(0)=-3,y^'(0)=-1 ?
The solution for y^{''}-6y^'+25y=0,y(0)=-3,y^'(0)=-1 is y=e^{3t}(-3cos(4t)+2sin(4t))