What is the solution for y^{''}-5y^'+6y=0,y(0)= 2/3 ?

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

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y^{''}-5y^'+6y=0,y(0)= 2/3

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = \frac{2}{3}

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

Solution steps

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

Solve linear ODE:

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

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

0.2 q^{^{''}} + 4 q^{^{'}} + 100 q = 0

y^{^{''}} - 20 y^{^{'}} + 100 y = 0, y (0) = 4, y (1) = 3

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 8 y^{^{'}} - 2 y = 0

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

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