IAL 2024 June Q4 | 國際數學站

(a) Show that the substitution $y^2 = \frac{1}{t}$ transforms the differential equation

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + y = xy^3 \qquad \text{(I)} \end{align*}

into the differential equation

\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} - 2t = -2x \qquad \text{(II)} \end{align*}

(3)

(b) Solve differential equation (II) and determine $y^2$ in terms of $x$ .

(6)

解答

(a)

Given $y^2 = \frac{1}{t}$ , we can write $y = t^{-1/2}$ . Differentiate with respect to $x$ using the chain rule:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,-\frac{1}{2}t^{-3/2} \frac{\mathrm{d}t}{\mathrm{d}x} \end{align*}

Substitute $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ into differential equation (I):

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + y = &\,xy^3\\[4mm] -\frac{1}{2}t^{-3/2} \frac{\mathrm{d}t}{\mathrm{d}x} + t^{-1/2} = &\,x(t^{-1/2})^3\\[4mm] -\frac{1}{2}t^{-3/2} \frac{\mathrm{d}t}{\mathrm{d}x} + t^{-1/2} = &\,xt^{-3/2} \end{align*}

Multiply the entire equation by $-2t^{3/2}$ :

\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} - 2t = &\,-2x \end{align*}

This is the required differential equation (II).

(b)

Differential equation (II) is a first-order linear ODE:

\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} - 2t = &\,-2x \end{align*}

The integrating factor (IF) is:

\begin{align*} \text{IF} = &\,\mathrm{e}^{\int -2 \,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{-2x} \end{align*}

Multiply the differential equation by the IF:

\begin{align*} \mathrm{e}^{-2x} \frac{\mathrm{d}t}{\mathrm{d}x} - 2\mathrm{e}^{-2x}t = &\,-2x\mathrm{e}^{-2x}\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\left( t\mathrm{e}^{-2x} \right) = &\,-2x\mathrm{e}^{-2x} \end{align*}

Integrate both sides with respect to $x$ . Let’s use integration by parts for the right hand side, where $u = x$ and $\mathrm{d}v = -2\mathrm{e}^{-2x} \,\mathrm{d}x$ , meaning $\mathrm{d}u = \mathrm{d}x$ and $v = \mathrm{e}^{-2x}$ :

\begin{align*} t\mathrm{e}^{-2x} = &\,\int -2x\mathrm{e}^{-2x} \,\mathrm{d}x\\[4mm] = &\,x\mathrm{e}^{-2x} - \int \mathrm{e}^{-2x} \,\mathrm{d}x\\[4mm] = &\,x\mathrm{e}^{-2x} + \frac{1}{2}\mathrm{e}^{-2x} + C \end{align*}

Multiply through by $\mathrm{e}^{2x}$ to solve for $t$ :

\begin{align*} t = &\,x + \frac{1}{2} + C\mathrm{e}^{2x} \end{align*}

We were given that $y^2 = \frac{1}{t}$ , which implies $t = \frac{1}{y^2}$ . Substituting this back:

\begin{align*} \frac{1}{y^2} = &\,x + \frac{1}{2} + C\mathrm{e}^{2x}\\[4mm] y^2 = &\,\frac{1}{x + \frac{1}{2} + C\mathrm{e}^{2x}} \end{align*}