What is the solution for 2y^{''}-5y^'-2y=0 ?

The solution for 2y^{''}-5y^'-2y=0 is y=c_{1}e^{((5+sqrt(41))t)/4}+c_{2}e^{((5-sqrt(41))t)/4}

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

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

y = c_{1} e^{\frac{( 5 + 41 ) t}{4}} + c_{2} e^{\frac{( 5 - 41 ) t}{4}}

Solution steps

2 y^{''} (t) - 5 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 5 + 41 ) t}{4}} + c_{2} e^{\frac{( 5 - 41 ) t}{4}}

y = c_{1} e^{\frac{( 5 + 41 ) t}{4}} + c_{2} e^{\frac{( 5 - 41 ) t}{4}}

y^{^{'''}} - 3 y^{^{''}} - 22 y^{^{'}} + 24 y = 0

y^{^{''}} - 5 y^{^{'}} = 0, y^{^{'}} (0) = 2

y^{^{''}} + (a + 2) y^{^{'}} + 4 y = 0

y^{^{''}} - 5 y^{^{'}} + 6.25 y = 0

3 y^{^{''}} - 5 y = 0

What is the solution for 2y^{''}-5y^'-2y=0 ?
The solution for 2y^{''}-5y^'-2y=0 is y=c_{1}e^{((5+sqrt(41))t)/4}+c_{2}e^{((5-sqrt(41))t)/4}