What is the solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 3/2 ?

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

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y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 3/2

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{2}

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

Solution steps

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

Solve linear ODE:

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

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

\frac{1}{2} y^{^{''}} - \frac{1}{3} y^{^{'}} - 2 y = 0

y^{^{''}} + y^{^{'}} + 9 y = 0

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

y^{^{''}} + 4 y^{^{'}} + y 3 = 0

y^{^{''}} + 36 y = 0, y (0) = cos (6 x)

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