What is the solution for 2y^{''}-4y^'-y=0 ?

The solution for 2y^{''}-4y^'-y=0 is y=c_{1}e^{((2+sqrt(6))t)/2}+c_{2}e^{((2-sqrt(6))t)/2}

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2y^{''}-4y^'-y=0

2 y^{^{''}} - 4 y^{^{'}} - y = 0

y = c_{1} e^{\frac{( 2 + 6 ) t}{2}} + c_{2} e^{\frac{( 2 - 6 ) t}{2}}

Solution steps

2 y^{''} (t) - 4 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 2 + 6 ) t}{2}} + c_{2} e^{\frac{( 2 - 6 ) t}{2}}

y = c_{1} e^{\frac{( 2 + 6 ) t}{2}} + c_{2} e^{\frac{( 2 - 6 ) t}{2}}

What is the solution for 2y^{''}-4y^'-y=0 ?
The solution for 2y^{''}-4y^'-y=0 is y=c_{1}e^{((2+sqrt(6))t)/2}+c_{2}e^{((2-sqrt(6))t)/2}