What is the solution for 2y^{''}-2y^'-15y=0 ?

The solution for 2y^{''}-2y^'-15y=0 is y=c_{1}e^{((1+sqrt(31))t)/2}+c_{2}e^{((1-sqrt(31))t)/2}

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2y^{''}-2y^'-15y=0

2 y^{^{''}} - 2 y^{^{'}} - 15 y = 0

y = c_{1} e^{\frac{( 1 + 31 ) t}{2}} + c_{2} e^{\frac{( 1 - 31 ) t}{2}}

Solution steps

2 y^{''} (t) - 2 y^{'} (t) - 15 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 31 ) t}{2}} + c_{2} e^{\frac{( 1 - 31 ) t}{2}}

y = c_{1} e^{\frac{( 1 + 31 ) t}{2}} + c_{2} e^{\frac{( 1 - 31 ) t}{2}}

4 y^{^{''}} - 784 y = 0, y (0) = 0

y^{^{'''}} + 5 y^{^{''}} - 16 y^{^{'}} - 170 y = 0

y^{^{''}} - 35 y = 0

y^{^{''''}} - 4 y^{^{'''}} - 16 y^{^{''}} + 64 y^{^{'}} = 0

5 y^{^{''}} - 7 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 0.8

What is the solution for 2y^{''}-2y^'-15y=0 ?
The solution for 2y^{''}-2y^'-15y=0 is y=c_{1}e^{((1+sqrt(31))t)/2}+c_{2}e^{((1-sqrt(31))t)/2}