IAL 2023 June Q4 | 國際數學站

(a) Determine the general solution of the differential equation

\begin{align*} \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 8\frac{\mathrm{d}y}{\mathrm{d}x} + 16y =\,& 48x^2 - 34\\[2mm] \end{align*}

(5)

Given that $y = 4$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 21$ at $x = 0$

(b) determine the particular solution of the differential equation.

(4)

(c) Hence find the value of $y$ at $x = -2$ , giving your answer in the form $p\mathrm{e}^q + r$ where $p$ , $q$ and $r$ are integers to be determined.

(2)

解答

(a)

To solve the differential equation $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 8\frac{\mathrm{d}y}{\mathrm{d}x} + 16y = 48x^2 - 34$ :

First, we find the complementary function (CF) by solving the auxiliary equation:

\begin{align*} m^2 - 8m + 16 = &\,0\\[4mm] (m-4)^2 = &\,0\\[4mm] m = &\,4 \text{ (repeated root)} \end{align*}

So the complementary function is $y_c = (A+Bx)\mathrm{e}^{4x}$ .

For the particular integral (PI), let $y_p = \lambda x^2 + \mu x + \nu$ . Differentiate with respect to $x$ :

\begin{align*} y_p' = &\,2\lambda x + \mu\\[4mm] y_p'' = &\,2\lambda \end{align*}

Substitute these into the differential equation:

\begin{align*} 2\lambda - 8(2\lambda x + \mu) + 16(\lambda x^2 + \mu x + \nu) = &\,48x^2 - 34\\[4mm] 16\lambda x^2 + (-16\lambda + 16\mu)x + (2\lambda - 8\mu + 16\nu) = &\,48x^2 - 34 \end{align*}

Equate the coefficients: For $x^2$ :

\begin{align*} 16\lambda = &\,48 \implies \lambda = 3 \end{align*}

For $x$ :

\begin{align*} -16\lambda + 16\mu = &\,0 \implies \mu = \lambda = 3 \end{align*}

For the constant term:

\begin{align*} 2\lambda - 8\mu + 16\nu = &\,-34\\[4mm] 2(3) - 8(3) + 16\nu = &\,-34\\[4mm] 6 - 24 + 16\nu = &\,-34\\[4mm] 16\nu = &\,-16 \implies \nu = -1 \end{align*}

The particular integral is $y_p = 3x^2 + 3x - 1$ . The general solution is:

\begin{align*} y = &\,(A+Bx)\mathrm{e}^{4x} + 3x^2 + 3x - 1 \end{align*}

(b)

Given that $y = 4$ at $x = 0$ :

\begin{align*} 4 = &\,(A+0)\mathrm{e}^0 + 3(0)^2 + 3(0) - 1\\[4mm] 4 = &\,A - 1\\[4mm] A = &\,5 \end{align*}

Now, differentiate the general solution to find $B$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,B\mathrm{e}^{4x} + 4(A+Bx)\mathrm{e}^{4x} + 6x + 3 \end{align*}

Given that $\frac{\mathrm{d}y}{\mathrm{d}x} = 21$ at $x = 0$ :

\begin{align*} 21 = &\,B\mathrm{e}^0 + 4(5+0)\mathrm{e}^0 + 6(0) + 3\\[4mm] 21 = &\,B + 20 + 3\\[4mm] 21 = &\,B + 23\\[4mm] B = &\,-2 \end{align*}

The particular solution is:

\begin{align*} y = &\,(5-2x)\mathrm{e}^{4x} + 3x^2 + 3x - 1 \end{align*}

(c)

Substitute $x = -2$ into the particular solution:

\begin{align*} y = &\,(5 - 2(-2))\mathrm{e}^{4(-2)} + 3(-2)^2 + 3(-2) - 1\\[4mm] = &\,(5 + 4)\mathrm{e}^{-8} + 3(4) - 6 - 1\\[4mm] = &\,9\mathrm{e}^{-8} + 12 - 7\\[4mm] = &\,9\mathrm{e}^{-8} + 5 \end{align*}

This is in the form $p\mathrm{e}^q + r$ , where $p=9$ , $q=-8$ , and $r=5$ .