What is the solution for 20y^{''}+3y^'-2y=0 ?

The solution for 20y^{''}+3y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-(2t)/5}

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20y^{''}+3y^'-2y=0

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

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

3 \frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 8 y = 0

x^{^{''}} - 6 x^{^{'}} + 10 x = 0

625 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

(D^{3} - D^{2} - 4 D + 4) y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y^{0} = 0, y^{^{'}} 0 = 1

What is the solution for 20y^{''}+3y^'-2y=0 ?
The solution for 20y^{''}+3y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-(2t)/5}