IAL 2025 June Q4 | 國際數學站

Given that $y = \lambda xe^{3x}$ is a particular integral of the differential equation

\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 11\frac{\mathrm{d}y}{\mathrm{d}x} - 3y = 78e^{3x} \end{align*}

(a) determine the value of the constant $\lambda$

(3)

(b) Hence determine the general solution of the differential equation.

(3)

Given also that $y = \dfrac{9}{2}$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ at $x = 0$

(4)

解答

(a)

Given the particular integral $y = \lambda x\mathrm{e}^{3x}$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\lambda \mathrm{e}^{3x} + 3\lambda x\mathrm{e}^{3x}\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,3\lambda \mathrm{e}^{3x} + 3\lambda \mathrm{e}^{3x} + 9\lambda x\mathrm{e}^{3x}\\[4mm] = &\,6\lambda \mathrm{e}^{3x} + 9\lambda x\mathrm{e}^{3x} \end{align*}

Substitute into the differential equation $4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 11\frac{\mathrm{d}y}{\mathrm{d}x} - 3y = 78\mathrm{e}^{3x}$ :

\begin{align*} &\,4(6\lambda \mathrm{e}^{3x} + 9\lambda x\mathrm{e}^{3x}) - 11(\lambda \mathrm{e}^{3x} + 3\lambda x\mathrm{e}^{3x}) - 3(\lambda x\mathrm{e}^{3x})\\[4mm] = &\,78\mathrm{e}^{3x} \end{align*}

Comparing coefficients of $\mathrm{e}^{3x}$ and $x\mathrm{e}^{3x}$ :

\begin{align*} x\mathrm{e}^{3x}: &\, 36\lambda - 33\lambda - 3\lambda = 0 \quad \text{(consistent)}\\[4mm] \mathrm{e}^{3x}: &\, 24\lambda - 11\lambda = 78\\[4mm] &\, 13\lambda = 78\\[4mm] &\, \lambda = 6 \end{align*}

(b)

To find the complementary function, solve the auxiliary equation:

\begin{align*} 4m^2 - 11m - 3 = &\,0\\[4mm] (4m + 1)(m - 3) = &\,0 \end{align*}

So $m = -\frac{1}{4}$ or $m = 3$ . The complementary function is $y_c = A\mathrm{e}^{-\frac{1}{4}x} + B\mathrm{e}^{3x}$ .

The general solution is:

\begin{align*} y = A\mathrm{e}^{-\frac{1}{4}x} + B\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} \end{align*}

(c)

Differentiating the general solution:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,-\frac{1}{4}A\mathrm{e}^{-\frac{1}{4}x} + 3B\mathrm{e}^{3x} + 6\mathrm{e}^{3x} + 18x\mathrm{e}^{3x} \end{align*}

At $x = 0$ , $y = \frac{9}{2}$ :

\begin{align*} \frac{9}{2} = A + B \end{align*}

At $x = 0$ , $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ :

\begin{align*} 0 = &\,-\frac{1}{4}A + 3B + 6\\[4mm] \frac{1}{4}A - 3B = &\,6\\[4mm] A - 12B = &\,24 \end{align*}

Solving the simultaneous equations:

\begin{align*} (A + B) - (A - 12B) = &\,\frac{9}{2} - 24\\[4mm] 13B = &\,-\frac{39}{2}\\[4mm] B = &\,-\frac{3}{2} \end{align*}

\begin{align*} A = &\,\frac{9}{2} - B\\[4mm] = &\,\frac{9}{2} - \left(-\frac{3}{2}\right)\\[4mm] = &\,6 \end{align*}

The particular solution is:

\begin{align*} y = 6\mathrm{e}^{-\frac{1}{4}x} - \frac{3}{2}\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} \end{align*}