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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{4} - 6 d^{3} - 12 d^{2} + 38 d - 21) y = 0

20 y^{^{''}} + 80 y^{^{'}} + 60 y = 0

\frac{d ^{2} y}{d t ^{2}} + 9 y = 0

y^{^{''}} + 81 y = 0, y (0) = 4, y^{^{'}} (0) = 1

4 y^{^{''}} - 12 y^{^{'}} + 10 y = 0

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