What is the solution for y^{''}-6y^'+8y=0,y(0)= 3/2 ,y^'(0)= 5/2 ?

The solution for y^{''}-6y^'+8y=0,y(0)= 3/2 ,y^'(0)= 5/2 is y=(3/2-7/4)e^{4t}+7/4 e^{2t}

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y^{''}-6y^'+8y=0,y(0)= 3/2 ,y^'(0)= 5/2

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = \frac{3}{2}, y^{^{'}} (0) = \frac{5}{2}

y = (\frac{3}{2} - \frac{7}{4}) e^{4 t} + \frac{7}{4} e^{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = (\frac{3}{2} - \frac{7}{4}) e^{4 t} + \frac{7}{4} e^{2 t}

y = (\frac{3}{2} - \frac{7}{4}) e^{4 t} + \frac{7}{4} e^{2 t}

y^{^{'''}} + 19 y^{^{''}} + 93 y^{^{'}} + 75 y = 0

2 y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 6 y = 0

y^{^{''}} + 10 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 0

16 y^{^{''}} - 8 y + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

x^{^{''}} + 12 x^{^{'}} + 40 x = 0

What is the solution for y^{''}-6y^'+8y=0,y(0)= 3/2 ,y^'(0)= 5/2 ?
The solution for y^{''}-6y^'+8y=0,y(0)= 3/2 ,y^'(0)= 5/2 is y=(3/2-7/4)e^{4t}+7/4 e^{2t}