What is the solution for y^{''}-12y^'+32y=0,y(0)=-1,y^'(0)=-2 ?

The solution for y^{''}-12y^'+32y=0,y(0)=-1,y^'(0)=-2 is y=(-1+3/2)e^{8t}-3/2 e^{4t}

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y^{''}-12y^'+32y=0,y(0)=-1,y^'(0)=-2

y^{^{''}} - 12 y^{^{'}} + 32 y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

y = (- 1 + \frac{3}{2}) e^{8 t} - \frac{3}{2} e^{4 t}

Solution steps

y^{''} (t) - 12 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = (- 1 + \frac{3}{2}) e^{8 t} - \frac{3}{2} e^{4 t}

y = (- 1 + \frac{3}{2}) e^{8 t} - \frac{3}{2} e^{4 t}

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 5, y (0) = 4

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y^{^{''}} + 16 y^{^{'}} + 3 y = 0

y^{^{''}} - 25 y^{^{'}} + 5 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

What is the solution for y^{''}-12y^'+32y=0,y(0)=-1,y^'(0)=-2 ?
The solution for y^{''}-12y^'+32y=0,y(0)=-1,y^'(0)=-2 is y=(-1+3/2)e^{8t}-3/2 e^{4t}