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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to - 5 - lim (\frac{3 x}{2 x + 10})

x \to 0 - lim ((\frac{1}{5})^{\frac{1}{x}})

x \to 1 - 0 lim (\frac{1}{1 - x})

x \to \infty + lim (x ln (x))

f (x) = x^{3} - 7 x^{2} + 6 x + 1

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