What is the solution for 2y^{''}-9y^'-5y=0 ?

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

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2y^{''}-9y^'-5y=0

2 y^{^{''}} - 9 y^{^{'}} - 5 y = 0

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

Solution steps

2 y^{''} (t) - 9 y^{'} (t) - 5 y = 0

Solve linear ODE:

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

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

sec^{2} (π x)

\frac{d}{d x} (\frac{6}{x + 2})

f (x) = (2 + 3 x) (6 - 5 x), at x = 1

\frac{d}{d x} (e^{(3 - x)})

\frac{d}{d x} (x e^{- 3 x})

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