IB Maths AA HL 5.18 Differential equations

[Maximum number: 9]

The following question explores features of a family of curves. The family is then linked to a homogeneous differential equation.
Consider the curve given by $y=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ .

(a)

Hence, show that a solution to the original differential equation may be expressed in the form $x^{2}=\left|\frac{A(x+y)}{x-y}\right|$ , where A is a positive constant.

Now consider only the case where $\frac{A(x+y)}{x-y}>0$ .

[ 5 ]

attempt to separate variables and integrate

\begin{aligned} & \int \frac{\mathrm{d} x}{x}=\int \frac{\mathrm{d} v}{1-v^{2}} \\ & \ln |x|=\frac{1}{2} \ln \left|\frac{A(1+v)}{1-v}\right| \end{aligned}

attempt to use power rule of logs

\begin{aligned} & \ln |x|=\ln \left|\frac{A(1+v)}{1-v}\right|^{\frac{1}{2}} \text { OR } \ln |x|^{2}=\ln \left|\frac{A(1+v)}{1-v}\right| \\ & |x|=\left|\frac{A(1+v)}{1-v}\right|^{\frac{1}{2}} \text { OR } x^{2}=\left|\frac{A(1+v)}{1-v}\right| \\ & x^{2}=\left|\frac{A\left(1+\frac{y}{x}\right)}{1-\frac{y}{x}}\right| \\ & x^{2}=\left|\frac{A(x+y)}{x-y}\right| \end{aligned}

Note: Award at most (M1)A1(M1)A1A0 for incorrect use or omission of moduli.
Note: The substitution of $\frac{y}{x}$ may be seen earlier.

(b)

Show that a solution to the original differential equation is $y=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$ .

[ 4 ]

$x^{2}=\frac{A(x+y)}{x-y}$
attempt to clear denominator
$x^{2}(x-y)=A(x+y)$ $x^{3}-x^{2} y=A x+A y$
attempt to isolate y
$x^{3}-A x=x^{2} y+A y$ $x\left(x^{2}-A\right)=y\left(x^{2}+A\right)\left(x^{3}-A x=y\left(x^{2}+A\right)\right)$ $y=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$

Note: The y must be factored out for the final A1.

IB Maths AA HL Ahl 5 18 Hl Differential Equations Questions

Exam points

IB Maths AA HL Ahl 5 18 Hl Differential Equations Questions question 1

Question (a)

Question (b)