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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3}), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} \cdot ln (9 + x^{2})

y^{^{''}} - 4 y = (x - 2) e^{x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 15

(D^{2} + 1) y = sec^{3} (x)

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