What is the solution for y^{''}-2y-3y^'=0 ?

The solution for y^{''}-2y-3y^'=0 is y=c_{1}e^{((3+sqrt(17))t)/2}+c_{2}e^{((3-sqrt(17))t)/2}

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y^{''}-2y-3y^'=0

y^{^{''}} - 2 y - 3 y^{^{'}} = 0

y = c_{1} e^{\frac{( 3 + 17 ) t}{2}} + c_{2} e^{\frac{( 3 - 17 ) t}{2}}

Solution steps

y^{''} (t) - 2 y - 3 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 3 + 17 ) t}{2}} + c_{2} e^{\frac{( 3 - 17 ) t}{2}}

y = c_{1} e^{\frac{( 3 + 17 ) t}{2}} + c_{2} e^{\frac{( 3 - 17 ) t}{2}}

y^{^{''}} + 10 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

15 y^{^{''}} - 16 y^{^{'}} + y = 0, y (0) = 15, y^{^{'}} (0) = \frac{1}{15}

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

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

u^{^{''}} - a^{2} H (t) = 0

What is the solution for y^{''}-2y-3y^'=0 ?
The solution for y^{''}-2y-3y^'=0 is y=c_{1}e^{((3+sqrt(17))t)/2}+c_{2}e^{((3-sqrt(17))t)/2}