What is the solution for y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)= 1/4 ?

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

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y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)= 1/4

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = \frac{1}{4}

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

Solution steps

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

Solve linear ODE:

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

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

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