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\displaystyley = (A+Bx)e^{\alpha x}
\displaystyle \frac{dy}{dx} = 0 \hspace{10} \frac{d^2y}{dx^2} = 0
\displaystyle \frac{dy}{dx} = a \hspace{10} \frac{d^2y}{dx^2} = 0
\displaystyle 2a = 3 \hspace{10} a = \frac{3}{2}
\displaystyle 3a + 2b = 5 \hspace{10} \frac{9}{2} + 2b = 5 \hspace{10} b = \frac{1}{4}
\displaystyle \frac{dy}{dx} = 2ke^{2x} \hspace{10} \frac{d^2y}{dx^2} = 4ke^{2x}
\displaystyle a = -\frac{1}{2} \hspace{10} b = \frac{3}{20}
\displaystyle 2r = 2 \hspace{10} r = \frac{1}{2}
\displaystyle 6 + 2q = 12 \hspace{10} q = 3