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

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

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

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

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

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

x^{^{''}} + \frac{2}{16} x^{^{'}} + 2 x = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} + 100 y = 0, y (0) = - \frac{2}{3}, y^{^{'}} (0) = 5

y^{^{'''}} + 6 y^{^{''}} + 8 y^{^{'}} = 0

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

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