What is the solution for x^{''}-6x^'+8x=0,x(0)=3 ?

The solution for x^{''}-6x^'+8x=0,x(0)=3 is x=(3-c_{2})e^{4t}+c_{2}e^{2t}

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x^{''}-6x^'+8x=0,x(0)=3

x^{^{''}} - 6 x^{^{'}} + 8 x = 0, x (0) = 3

x = (3 - c_{2}) e^{4 t} + c_{2} e^{2 t}

Solution steps

x^{''} (t) - 6 x^{'} (t) + 8 x = 0

Solve linear ODE:

x = (3 - c_{2}) e^{4 t} + c_{2} e^{2 t}

x = (3 - c_{2}) e^{4 t} + c_{2} e^{2 t}

y^{^{''}} + y - 6 y^{^{'}} = 0

y^{^{''}} - 8 y^{^{'}} - 7 y = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (0) = 1

x^{^{''}} + 6 x^{^{'}} + 9 x = 0, x (0) = a, x^{^{'}} (0) = b

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

What is the solution for x^{''}-6x^'+8x=0,x(0)=3 ?
The solution for x^{''}-6x^'+8x=0,x(0)=3 is x=(3-c_{2})e^{4t}+c_{2}e^{2t}