What is the solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)= 1/2 ?

The solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)= 1/2 is y=(2-5/2)e^{(3t)/2}+5/2 e^{t/2}

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4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)= 1/2

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{1}{2}

y = (2 - \frac{5}{2}) e^{\frac{3 t}{2}} + \frac{5}{2} e^{\frac{t}{2}}

Solution steps

4 y^{''} (t) - 8 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = (2 - \frac{5}{2}) e^{\frac{3 t}{2}} + \frac{5}{2} e^{\frac{t}{2}}

y = (2 - \frac{5}{2}) e^{\frac{3 t}{2}} + \frac{5}{2} e^{\frac{t}{2}}

What is the solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)= 1/2 ?
The solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)= 1/2 is y=(2-5/2)e^{(3t)/2}+5/2 e^{t/2}