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

The solution for 4y^{''}-8y^'+3=0,y(0)=2,y^'(0)= 1/2 is y=(2-31/16)e^{2t}+31/16+(3t)/8

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

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

y = (2 - \frac{31}{16}) e^{2 t} + \frac{31}{16} + \frac{3 t}{8}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{31}{16}) e^{2 t} + \frac{31}{16} + \frac{3 t}{8}

y = (2 - \frac{31}{16}) e^{2 t} + \frac{31}{16} + \frac{3 t}{8}

What is the solution for 4y^{''}-8y^'+3=0,y(0)=2,y^'(0)= 1/2 ?
The solution for 4y^{''}-8y^'+3=0,y(0)=2,y^'(0)= 1/2 is y=(2-31/16)e^{2t}+31/16+(3t)/8