IAL 2025 Jan Q6 | 國際數學站

(a) Determine the general solution of the differential equation

\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\frac{\mathrm{d}y}{\mathrm{d}x} + 37y = 6\mathrm{e}^{5x} \end{align*}

(6)

Given that $y = 0$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$

(b) determine the particular solution for this differential equation.

(5)

解答

(a)

The auxiliary equation is:

\begin{align*} 4m^2 - 4m + 37 = &\,0\\[4mm] m = &\,\frac{4 \pm \sqrt{16 - 4(4)(37)}}{8}\\[4mm] = &\,\frac{4 \pm \sqrt{16 - 592}}{8}\\[4mm] = &\,\frac{4 \pm \sqrt{-576}}{8}\\[4mm] = &\,\frac{4 \pm 24\mathrm{i}}{8}\\[4mm] = &\,\frac{1}{2} \pm 3\mathrm{i} \end{align*}

The complementary function is:

\begin{align*} y_c = &\,\mathrm{e}^{\frac{1}{2}x}(A\cos 3x + B\sin 3x) \end{align*}

For the particular integral, try $y_p = k\mathrm{e}^{5x}$ :

\begin{align*} \frac{\mathrm{d}y_p}{\mathrm{d}x} = &\,5k\mathrm{e}^{5x}\\[4mm] \frac{\mathrm{d}^2y_p}{\mathrm{d}x^2} = &\,25k\mathrm{e}^{5x} \end{align*}

Substituting into the differential equation:

\begin{align*} 4(25k\mathrm{e}^{5x}) - 4(5k\mathrm{e}^{5x}) + 37(k\mathrm{e}^{5x}) = &\,6\mathrm{e}^{5x}\\[4mm] 100k - 20k + 37k = &\,6\\[4mm] 117k = &\,6\\[4mm] k = &\,\frac{6}{117} = \frac{2}{39} \end{align*}

Thus, the general solution is:

\begin{align*} y = &\,\frac{2}{39}\mathrm{e}^{5x} + \mathrm{e}^{\frac{1}{2}x}(A\cos 3x + B\sin 3x) \end{align*}

(b)

Using the initial condition $y = 0$ when $x = 0$ :

\begin{align*} 0 = &\,\frac{2}{39}\mathrm{e}^0 + \mathrm{e}^0(A\cos 0 + B\sin 0)\\[4mm] 0 = &\,\frac{2}{39} + A\\[4mm] A = &\,-\frac{2}{39} \end{align*}

Differentiating the general solution with respect to $x$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{10}{39}\mathrm{e}^{5x} + \frac{1}{2}\mathrm{e}^{\frac{1}{2}x}(A\cos 3x + B\sin 3x) + \mathrm{e}^{\frac{1}{2}x}(-3A\sin 3x + 3B\cos 3x) \end{align*}

Using the initial condition $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$ :

\begin{align*} 0 = &\,\frac{10}{39}\mathrm{e}^0 + \frac{1}{2}\mathrm{e}^0(A\cos 0 + B\sin 0) + \mathrm{e}^0(-3A\sin 0 + 3B\cos 0)\\[4mm] 0 = &\,\frac{10}{39} + \frac{1}{2}A + 3B\\[4mm] 0 = &\,\frac{10}{39} + \frac{1}{2}\left(-\frac{2}{39}\right) + 3B\\[4mm] 0 = &\,\frac{10}{39} - \frac{1}{39} + 3B\\[4mm] 3B = &\,-\frac{9}{39}\\[4mm] B = &\,-\frac{3}{39} = -\frac{1}{13} \end{align*}

So the particular solution is:

\begin{align*} y = &\,\frac{2}{39}\mathrm{e}^{5x} + \mathrm{e}^{\frac{1}{2}x}\left(-\frac{2}{39}\cos 3x - \frac{1}{13}\sin 3x\right) \end{align*}