What is the solution for y^{''}-5y^'+6y=0,y(0)=-1,y^'(0)=-7/2 ?

The solution for y^{''}-5y^'+6y=0,y(0)=-1,y^'(0)=-7/2 is y=(-1-1/2)e^{3t}+1/2 e^{2t}

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y^{''}-5y^'+6y=0,y(0)=-1,y^'(0)=-7/2

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = - \frac{7}{2}

y = (- 1 - \frac{1}{2}) e^{3 t} + \frac{1}{2} e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = (- 1 - \frac{1}{2}) e^{3 t} + \frac{1}{2} e^{2 t}

y = (- 1 - \frac{1}{2}) e^{3 t} + \frac{1}{2} e^{2 t}

y^{^{''''}} + 18 y^{^{''}} - 81 y = 0

y^{^{''}} + 10 y^{^{'}} + 288 y = 0

(d^{2} + 6) y = 0

y^{^{''}} + 9.82 \cdot y^{^{'}} = 0

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

What is the solution for y^{''}-5y^'+6y=0,y(0)=-1,y^'(0)=-7/2 ?
The solution for y^{''}-5y^'+6y=0,y(0)=-1,y^'(0)=-7/2 is y=(-1-1/2)e^{3t}+1/2 e^{2t}