What is the solution for 2y^{''}+5y^'-18y=0 ?

The solution for 2y^{''}+5y^'-18y=0 is y=c_{1}e^{2t}+c_{2}e^{-(9t)/2}

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2y^{''}+5y^'-18y=0

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

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

x \to - 1 lim (\frac{1 + x ^{3}}{1 + x})

\frac{d}{d x} (6 x^{- \frac{1}{2}})

y = 2 x^{2} + \frac{3}{x}

\frac{d}{d t} (0.1 cos (1 0^{5}) t)

What is the solution for 2y^{''}+5y^'-18y=0 ?
The solution for 2y^{''}+5y^'-18y=0 is y=c_{1}e^{2t}+c_{2}e^{-(9t)/2}