IAL 2023 June Q7 | 國際數學站

(a) Show that the substitution $z = y^{-2}$ transforms the differential equation

\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} + y + 4x^2y^3\ln x =\,& 0 \qquad x > 0 \qquad \text{(I)}\\[2mm] \end{align*}

into the differential equation

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{2z}{x} =\,& 8x\ln x \qquad x > 0 \qquad \text{(II)}\\[2mm] \end{align*}

(5)

(b) By solving differential equation (II), determine the general solution of differential equation (I), giving your answer in the form $y^2 = \mathrm{f}(x)$

(6)

解答

(a)

Given $z = y^{-2}$ , differentiate with respect to $x$ using the chain rule:

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

We are given differential equation (I):

\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} + y + 4x^2y^3\ln x = &\,0 \end{align*}

Divide the entire equation by $y^3$ :

\begin{align*} x y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} + y^{-2} + 4x^2\ln x = &\,0 \end{align*}

Substitute $y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}\frac{\mathrm{d}z}{\mathrm{d}x}$ and $y^{-2} = z$ :

\begin{align*} x \left(-\frac{1}{2}\frac{\mathrm{d}z}{\mathrm{d}x}\right) + z + 4x^2\ln x = &\,0\\[4mm] -\frac{x}{2}\frac{\mathrm{d}z}{\mathrm{d}x} + z = &\,-4x^2\ln x \end{align*}

Multiply the equation by $-\frac{2}{x}$ (since $x>0$ ):

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{2}{x}z = &\,8x\ln x \end{align*}

This matches differential equation (II).

(b)

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

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{2}{x}z = &\,8x\ln x \end{align*}

Find the integrating factor (IF):

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

Multiply (II) by the IF:

\begin{align*} x^{-2}\frac{\mathrm{d}z}{\mathrm{d}x} - 2x^{-3}z = &\,x^{-2}(8x\ln x)\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}(x^{-2}z) = &\,8x^{-1}\ln x \end{align*}

Integrate both sides with respect to $x$ :

\begin{align*} x^{-2}z = &\,\int 8x^{-1}\ln x \,\mathrm{d}x \end{align*}

Using the substitution $u = \ln x \implies \mathrm{d}u = x^{-1} \mathrm{d}x$ :

\begin{align*} \int 8x^{-1}\ln x \,\mathrm{d}x = &\,\int 8u \,\mathrm{d}u\\[4mm] = &\,4u^2 + k\\[4mm] = &\,4(\ln x)^2 + k \end{align*}

So we have:

\begin{align*} x^{-2}z = &\,4(\ln x)^2 + k\\[4mm] z = &\,4x^2(\ln x)^2 + kx^2 \end{align*}

We were given that $z = y^{-2}$ , so $z = \frac{1}{y^2}$ . Substituting this back gives the general solution for $y$ :

\begin{align*} \frac{1}{y^2} = &\,4x^2(\ln x)^2 + kx^2\\[4mm] y^2 = &\,\frac{1}{4x^2(\ln x)^2 + kx^2} \quad \text{or} \quad \frac{1}{x^2(4(\ln x)^2 + k)} \end{align*}