What is the solution for 6y^{''}-y^'-y=0,y(0)=1.25 ?

The solution for 6y^{''}-y^'-y=0,y(0)=1.25 is y=(1.25-c_{2})e^{t/2}+c_{2}e^{-t/3}

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6y^{''}-y^'-y=0,y(0)=1.25

6 y^{^{''}} - y^{^{'}} - y = 0, y (0) = 1.25

y = (1.25 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{3}}

Solution steps

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

Solve linear ODE:

y = (1.25 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{3}}

y = (1.25 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{3}}

y^{^{''}} - 11 y + 24 y = 0

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = α, y^{^{'}} (0) = 1

4 y^{^{''}} + 8 y^{^{'}} - 21 y = 0, y (0) = 1, y^{^{'}} (0) = 0

u^{^{''}} + 7 H (t) + 2 H (t) = 0

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = α, y^{^{'}} (0) = 5

What is the solution for 6y^{''}-y^'-y=0,y(0)=1.25 ?
The solution for 6y^{''}-y^'-y=0,y(0)=1.25 is y=(1.25-c_{2})e^{t/2}+c_{2}e^{-t/3}