Tangent Tuition
Specialist Mathematics · Past QCAA questions All 24 questions
2023 · Paper 1 · Technology-free

Question 18

[6 marks] Technology-free
Unit 4 · Topic 3Rates of change and differential equations Tool Kit 7.3Separable differential equations Tool Kit 6.2Substitution (step 2)
The question

A particular solution to the differential equation $\dfrac{dy}{dx}=\dfrac{x}{\left(x^{2}+1\right)\tan(y)}$, where $x\ge0$ and $-\dfrac{\pi}{2}<y\le0$, passes through the origin.

Determine this solution in the form $x=f(y)$. Leave your answer in simplified form.

Watch the situation first

Two curves fit the equation, but only one fits the conditions

Watch the solution follow the slope field out of the origin. You will see why squaring gives you a second curve that you have to throw away.

The slope depends on both x and y The conditions keep you in the shaded strip The solution leaves the origin along the slopes Squaring gives you two branches, x = ±tan(y) Only x = −tan(y) stays in the strip y = −π2 x y 1 2 3 x = tan(y) x = −tan(y) dydx = x(1 + x²) tan y each dash shows the slope x ≥ 0 −π2 < y ≤ 0 ∫ tan(y) dy = ∫ x1 + x² dx −ln(cos y) = ½ ln(1 + x²) 1 + x² = sec²(y) x² = tan²(y), so x = ±tan(y) y ≤ 0 makes tan(y) ≤ 0, but x ≥ 0 x = −tan(y)

The white curve also solves the differential equation, which is why you need the conditions to choose between them.

Now for the mathematics

Change the constant until the curve passes through the origin

Every value of $c$ in $-\ln(\cos y)=\frac{1}{2}\ln\left(1+x^{2}\right)+c$ gives you a different curve. You need the one that goes through $(0,\ 0)$.

y = −π2 x y your curve x = −tan(y)
Where the curve starts
Solution for this c
from 1 + x² = e^(−2c) sec²(y)
Through the origin?

QCAA marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. You can follow through from an earlier slip, so keep going even if you are unsure of one line.

Step 11 mark

You put every $y$ with $dy$ and every $x$ with $dx$. Multiply both sides by $\tan(y)$ to move it across.

$$\int\tan(y)\,dy=\int\frac{x}{x^{2}+1}\,dx$$

This mark can be implied by your later working.

Marker: correctly separates the variables

Step 21 mark

Both sides are the derivative of something over that something, as long as you fix the constants. Write $\tan(y)$ as $\frac{\sin(y)}{\cos(y)}$ and put a 2 on top of the $x$ side.

$$-\int\frac{-\sin(y)}{\cos(y)}\,dy=\frac{1}{2}\int\frac{2x}{x^{2}+1}\,dx$$

$$-\ln\left(\cos(y)\right)=\frac{1}{2}\ln\left(x^{2}+1\right)+c$$

You do not need absolute value signs here, because $\cos(y)>0$ for $-\frac{\pi}{2}<y\le0$ and $x^{2}+1>0$. Follow-through marks are allowed.

Marker: applies suitable integration methods

Step 31 mark

The curve passes through the origin, so you substitute $x=0$ and $y=0$.

$$-\ln\left(\cos(0)\right)=\frac{1}{2}\ln(1)+c\quad\Rightarrow\quad0=0+c\quad\Rightarrow\quad c=0$$

You have to show this substitution to earn the mark, even though the answer is 0. Your constant might look different if you wrote the general solution another way.

Marker: determines a value for the constant of integration

Step 4 · insight mark ★1 mark

Now you get rid of the logarithms. Use log laws to move the $\frac{1}{2}$ and the minus sign inside, so both sides are a single log.

$$\ln\left(\left(1+x^{2}\right)^{\frac{1}{2}}\right)=\ln\left(\frac{1}{\cos(y)}\right)$$

$$\left(1+x^{2}\right)^{\frac{1}{2}}=\sec(y)\quad\Rightarrow\quad1+x^{2}=\sec^{2}(y)$$

You can also write $1+x^{2}=\frac{1}{\cos^{2}(y)}$, and it is just as good.

Marker: determines an expression for a solution that does not contain logarithms

Step 51 mark

You take 1 from both sides and use the identity $\sec^{2}(y)-1=\tan^{2}(y)$.

$$x^{2}=\sec^{2}(y)-1=\tan^{2}(y)\quad\Rightarrow\quad x=\pm\tan(y)$$

Stopping at $x=\pm\sqrt{\sec^{2}(y)-1}$ also earns you this mark. This mark can be implied by your later working.

Marker: expresses $x$ in terms of $y$

Step 61 mark

Now you use the conditions to choose the sign. For $-\frac{\pi}{2}<y\le0$, $\tan(y)\le0$. You also need $x\ge0$, so you have to take the negative.

$x=-\tan(y)$

Say in words why you rejected $x=\tan(y)$, because that reasoning is part of this mark. The animation above shows that $x=\tan(y)$ is a real solution of the differential equation, but it sits outside the strip you are allowed to use.

Marker: evaluates the reasonableness of the results and expresses the solution in the form of $x=f(y)$ in simplified form

Putting it all together

Each phrase of the question gave you something

“$\frac{dy}{dx}=\frac{x}{\left(1+x^{2}\right)\tan(y)}$”
You can separate it, because it is a function of $x$ times a function of $y$.
“where $x\ge0$ and $-\frac{\pi}{2}<y\le0$”
You can drop the absolute values, and you use these conditions again to choose the sign at the end.
“passes through the origin”
You substitute $(0,\ 0)$ to find $c=0$.
“in the form $x=f(y)$ … in simplified form”
You make $x$ the subject and simplify $\sqrt{\sec^{2}(y)-1}$ to $-\tan(y)$.
What makes this complex unfamiliar

Separating and integrating is familiar. What makes this one harder is that you are asked for $x$ in terms of $y$, so you have to undo two logarithms and a trig identity, and then decide between two signs.

When I mark this kind of question, the line I look for first is the one where you choose the sign. The QCAA like to put conditions on $x$ and $y$ in the stem, and you should expect to use them before you finish.

Only mark this done when you could do it without help. Reading the solution does not count.

Back to the index

Question wording and marking-guide steps are from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.