Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 1 · Simple familiar

Question 13

[6 marks] Technology-free
The question

A slope field for the differential equation $\dfrac{dy}{dx}=xy$ is shown. Line segments have not been drawn at the three points marked with a dot.

−2 −2 −1 −1 1 1 2 2 O x y
a)

Complete the slope field by sketching the slopes at the three points indicated.

[1 mark]
b)

Use the completed slope field to sketch the solution curve for $\dfrac{dy}{dx}=xy$, given the point $(0,\,1)$ lies on this curve.

[1 mark]
c)

Use separation of variables to show that the general solution of the differential equation is $y=Ae^{\frac{x^{2}}{2}}$, where $A$ is a constant.

[3 marks]
d)

Determine the equation of the solution curve sketched in Question 13b).

[1 mark]
Watch the situation first

Fill the gaps, then let the field carry the curve

Each segment shows the gradient at its point. Watch the three missing ones go in, then follow the field from $(0,\,1)$ in both directions.

−2 −2 −1 −1 1 1 2 2 O x y y = ex²/2 Gradient at (x, y) = xy (1, 1): (1)(1) = 1 (−1, 1): (−1)(1) = −1 (2, −1): (2)(−1) = −2 (0, 1): (0)(1) = 0, flat At x = 1, y = √e ≈ 1.65 gradient ≈ 1.65, steeper than 1 Lowest point (0, 1) y ≥ 1: never reaches the x-axis A slope field for dy/dx = xyThe segment at (x, y) has gradient xyThree segments are missing: multiply the coordinatesThe curve starts at (0, 1), where the gradient is 0Follow the segments: the curve keeps getting steeperThe field is symmetrical about the y-axis, so the curve is tooOne smooth U shape, lowest at (0, 1), above the x-axis

Drawn to scale on the printed grid. The orange tangent always lies along the field, which is what makes the curve a solution.

Now for the mathematics

Draw the missing segments, then pick the curve

For part a), pick a dot and set the gradient of your segment. For parts b) to d), choose $A$ in $y=Ae^{\frac{x^{2}}{2}}$ until the curve passes through $(0,\,1)$, then move along it and watch its tangent follow the field.

−2 −2 −1 −1 1 1 2 2 O x y
Your three segments Tangent to the curve
Choose a part
Choose a dot

Marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

Substitute each marked point into $\dfrac{dy}{dx}=xy$, then draw a short segment with that gradient through each dot.

$$\begin{aligned}(1,\,1)\text{:}\quad\frac{dy}{dx}&=(1)(1)=1\\(-1,\,1)\text{:}\quad\frac{dy}{dx}&=(-1)(1)=-1\\(2,\,-1)\text{:}\quad\frac{dy}{dx}&=(2)(-1)=-2\end{aligned}$$

Segments drawn with gradients $1$, $-1$ and $-2$
Marking

Each segment must have the correct sign and an approximately correct gradient.

Marker: correctly draws all three line segments

Step 2 · part b)1 mark

Start at $(0,\,1)$, where the gradient is $0$, and follow the segments out to each side.

Draw a curve through $(0,\,1)$ with a minimum turning point there, symmetrical about the $y$-axis, decreasing for $x<0$ and increasing for $x>0$, becoming steep for $\lvert x\rvert>1$ and following the slope field.

−2 −2 −1 −1 1 1 2 2 O x y
Marking

The curve must pass through $(0,\,1)$ and must not cross the $x$-axis.

Marker: correctly sketches the solution curve

Step 3 · part c)1 mark

Divide both sides by $y$ and integrate, so each variable sits with its own differential.

$$\begin{aligned}\frac{1}{y}\,\frac{dy}{dx}&=x\\\int\frac{1}{y}\,dy&=\int x\,dx\end{aligned}$$

Marker: correctly separates the variables

Step 4 · part c)1 mark

Integrate each side. One constant on the right is enough.

$$\ln\lvert y\rvert=\frac{x^{2}}{2}+c$$

Why?

The absolute value is needed because $y$ may be negative, and $\ln$ is only defined for positive inputs.

Marker: correctly integrates both sides

Step 5 · part c)1 mark

Raise $e$ to the power of each side, split off the constant and remove the absolute value.

$$\begin{aligned}\lvert y\rvert&=e^{c}\,e^{\frac{x^{2}}{2}}\\y&=Ae^{\frac{x^{2}}{2}},\quad\text{where } A=\pm e^{c}\end{aligned}$$

Why?

$\lvert y\rvert=k$ means $y=k$ or $y=-k$. Since $c$ is a constant, $\pm e^{c}$ is also a constant, so call it $A$.

$y=Ae^{\frac{x^{2}}{2}}$, as required

Marker: correctly rearranges to show the required result

Step 6 · part d)1 mark

Substitute $x=0$, $y=1$ into the general solution to find $A$.

$$\begin{aligned}1&=Ae^{0}\\A&=1\end{aligned}$$

$y=e^{\frac{x^{2}}{2}}$

Marker: correctly determines the equation of the solution curve

Putting it all together

Each phrase of the question gave you something

“A slope field for the differential equation $\dfrac{dy}{dx}=xy$”
Each segment’s gradient is $xy$, so the gradient at any point is its two coordinates multiplied.
“the three points marked with a dot”
Gradients $1$ at $(1,\,1)$, $-1$ at $(-1,\,1)$ and $-2$ at $(2,\,-1)$, each drawn with the right sign and steepness.
“given the point $(0,\,1)$ lies on this curve”
The curve is flat there, since $(0)(1)=0$, and $(0,\,1)$ is a minimum turning point. The curve is symmetrical about the $y$-axis and stays at $y\ge1$.
“Use separation of variables”
Divide by $y$ and integrate: $\displaystyle\int\frac{1}{y}\,dy=\int x\,dx$, so $\ln\lvert y\rvert=\dfrac{x^{2}}{2}+c$.
“where $A$ is a constant”
Split $e^{\frac{x^{2}}{2}+c}=e^{c}\,e^{\frac{x^{2}}{2}}$ and call $\pm e^{c}$ by the name $A$.
“the solution curve sketched in Question 13b)”
It passes through $(0,\,1)$, so $A=1$ and $y=e^{\frac{x^{2}}{2}}$.
Common errors

If I had to guess where you might lose marks…

  • Rise and run swapped in a). At $(2,\,-1)$ the gradient is $-2$, down 2 for every 1 across. Drawing down 1 for every 2 across, or misreading the equation as $\dfrac{dy}{dx}=\dfrac{y}{x}$, gives $-\dfrac{1}{2}$: a shallow segment (about $27^\circ$) instead of a steep one (about $63^\circ$). A sign slip at $(-1,\,1)$, drawing $+1$ instead of $-1$, loses the mark in the same way.
  • Curve drawn through the lattice points in b). Joining $(-1,\,1)$, $(0,\,1)$ and $(1,\,1)$ is flat for too long: the true curve passes about $0.65$ above those dots ($y=\sqrt{e}\approx1.65$ at $x=\pm1$). A V-shaped corner at $(0,\,1)$ or a curve that dips to the $x$-axis also loses the mark.
  • Missing the $\tfrac{1}{2}$ in $\displaystyle\int x\,dx$. Writing $\ln\lvert y\rvert=x^{2}+c$ leads to $y=Ae^{x^{2}}$ and, in d), $y=e^{x^{2}}$. Its derivative is $2xe^{x^{2}}=2xy$, not $xy$.
  • Adding the constant instead of multiplying. $e^{\frac{x^{2}}{2}+c}=e^{\frac{x^{2}}{2}}+e^{c}$ is false: for example $e^{\frac{1}{2}+1}\approx4.48$ but $e^{\frac{1}{2}}+e^{1}\approx4.37$. It gives the wrong family $y=e^{\frac{x^{2}}{2}}+c$, which happens to give the right curve in d), since $c=0$ through $(0,\,1)$, but loses the third mark in c).
  • Integrating without separating. Writing $\displaystyle\int xy\,dx=\frac{x^{2}y}{2}$ treats $y$ as a constant when it is a function of $x$. That step is invalid and earns no marks in c).

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

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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