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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

\frac{d y}{d x} = \frac{4 x ^{4} y + 6 y ^{5}}{2 x ^{5} + 3 x y ^{4}}

\int_{\frac{1}{2}}^{3} 6 x ln (2 x) d x

f (x) = 2 x - 3

\frac{d ^{2}}{d x ^{2}} (7 x^{\frac{3}{2}})

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