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

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

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

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

y = \frac{8}{3} e^{\frac{3 t}{8}} - \frac{8}{3}

Solution steps

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

Solve linear ODE:

y = \frac{8}{3} e^{\frac{3 t}{8}} - \frac{8}{3}

y = \frac{8}{3} e^{\frac{3 t}{8}} - \frac{8}{3}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 4, y^{^{'}} (0) = 16

(D^{2} - 4 D) y = 0

5 y^{^{''}} + 24 y^{^{'}} - 5 y = 0

y^{^{''}} - 14 y^{^{'}} + 58 y = 0, y (0) = 5, y^{^{'}} (0) = - 4

x^{^{''}} + 8 x^{^{'}} + 25 x = 0

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