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

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

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

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

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

Solution steps

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

x^{5} - x^{3}

y^{^{'}} = \frac{4 x}{y}

x \to 1 lim ((x) (x - 1)^{- 1})

\int_{0}^{1} x^{3} (1 - x)^{2} d x

\int x e^{- 10 x} d x

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