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}}

\frac{d y}{d x} = 4 e^{x - y}

(x y + y^{2} + y) d x + (x + 2 y) d y = 0

y^{^{'}} = cos (x + y)

\frac{\partial}{\partial x} (e^{x} cos (x + y))

sec (\frac{π}{3})

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}