IBDP Maths AA: Topic: AHL 5.18: First order differential equations: IB style Questions HL Paper 2

Question 

Consider the differential equation $\frac{d y}{d x}=\frac{4 x^2+y^2-x y}{x^2}$, with $y=2$ when $x=1$.
a. Use Euler’s method, with step length $h=0.1$, to find an approximate value of $y$ when $x=1.4$.
b. Sketch the isoclines for $\frac{d y}{d x}=4$.
c.i. Express $m^2-2 m+4$ in the form $(m-a)^2+b$, where $a, b \in Z$.
c.ii.Solve the differential equation, for $x>0$, giving your answer in the form $y=f(x)$.
c.iiisketch the graph of $y=f(x)$ for $1 \leqslant x \leqslant 1.4$.
c.ivWith reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f(1.4)$ will be less than, equal to, or greater than your answer in part (a).
▶️Answer/Explanation

 

Markscheme
a.
$$
y(1.4) \approx 5.34
$$
Note: Award $\boldsymbol{A} 1$ for each correct $y$ value.
For the intermediate $y$ values, accept answers that are accurate to 2 significant figures. The final $y$ value must be accurate to 3 significant figures or better.
[5 marks]
b. attempt to solve $\frac{4 x^2+y^2-x y}{x^2}=4$
(M1)
$$
\begin{aligned}
& \Rightarrow y^2-x y=0 \\
& y(y-x)=0 \\
& y=0 \text { or } y=x
\end{aligned}
$$
c.i. $m^2-2 m+4=(m-1)^2+3 \quad(a=1, b=3)$
A1
[1 mark]
c.ii.recognition of homogeneous equation,
let $y=v x \quad M 1$
the equation can be written as
$$
\begin{aligned}
& v+x \frac{\mathrm{d} v}{\mathrm{~d} x}=4+v^2-v \quad \text { (A1) } \\
& x \frac{\mathrm{d} v}{\mathrm{~d} x}=v^2-2 v+4 \\
& \int \frac{1}{v^2-2 v+4} \mathrm{~d} v=\int \frac{1}{x} \mathrm{~d} x \quad \text { M1 }
\end{aligned}
$$
$M 1$
Note: Award $\boldsymbol{M 1}$ for attempt to separate the variables.
$\int \frac{1}{(v-1)^2+3} \mathrm{~d} v=\int \frac{1}{x} \mathrm{~d} x$ from part (c)(i) $\quad$ M1
$\frac{1}{\sqrt{3}} \arctan \left(\frac{v-1}{\sqrt{3}}\right)=\ln x(+c) \quad$ A1A1
$$
x=1, y=2 \Rightarrow v=2
$$
$$
\frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right)=\ln 1+c \quad \text { M1 }
$$
Note: Award $\boldsymbol{M} 1$ for using initial conditions to find $c$.
$$
\begin{aligned}
& \Rightarrow c=\frac{\pi}{6 \sqrt{3}}(=0.302) \quad \text { A1 } \\
& \arctan \left(\frac{v-1}{\sqrt{3}}\right)=\sqrt{3} \ln x+\frac{\pi}{6} \\
& \text { substituting } v=\frac{y}{x} \quad \text { M1 }
\end{aligned}
$$
Note: This $\boldsymbol{M} 1$ may be awarded earlier.
$$
y=x\left(\sqrt{3} \tan \left(\sqrt{3} \ln x+\frac{\pi}{6}\right)+1\right) \quad \boldsymbol{A 1}
$$
c.iii.
curve drawn over correct domain
A1
[1 mark]
c.ivthe sketch shows that $f$ is concave up $\boldsymbol{A 1}$
Note: Accept $f^{\prime}$ is increasing.
this means the tangent drawn using Euler’s method will give an underestimate of the real value, so $f(1.4)>$ estimate in part (a)
Note: The $\boldsymbol{R} 1$ is dependent on the $\boldsymbol{A} 1$.
 
 

Question

The curve $y=f(x)$ has a gradient function given by
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=x-y
$$

The curve passes through the point $(1,1)$.
a.i. On the same set of axes, sketch and label isoclines for $\frac{\mathrm{d} y}{\mathrm{~d} x}=-1,0$ and 1 , and clearly indicate the value of each $y$-intercept.
a.ii.Hence or otherwise, explain why the point $(1,1)$ is a local minimum.
b. Find the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=x-y$, which passes through the point $(1,1)$. Give your answer in the form $y=f(x)$.
c.i. Explain why the graph of $y=f(x)$ does not intersect the isocline $\frac{\mathrm{d} y}{\mathrm{~d} x}=1$.
c.ii.Sketch the graph of $y=f(x)$ on the same set of axes as part (a)(i).

▶️Answer/Explanation

a.i.attempt to find equation of isoclines by setting $x-y=-1,0,1$

3 parallel lines with positive gradient
A1
$y$-intercept $=-c$ for $\frac{\mathrm{d} y}{\mathrm{~d} x}=c$
A1
Note: To award A1, each $y$-intercept should be clear, but condone a missing label (eg. $(0,0)$ ).
If candidates represent the lines using slope fields, but omit the lines, award maximum of M1AOA1.
[3 marks]
a.ii.at point $(1,1), \frac{\mathrm{d} y}{\mathrm{~d} x}=0$
A1
EITHER
to the left of $(1,1)$, the gradient is negative $\quad \boldsymbol{R} \boldsymbol{1}$
to the right of $(1,1)$, the gradient is positive $\quad \boldsymbol{R} 1$

Note: Accept any correct reasoning using gradient, isoclines or slope field.
If a candidate uses left/right or $x<1 / x>1$ without explicitly referring to the point $(1,1)$ or a correct region on the diagram, award $\boldsymbol{R O R 1}$.

OR
$$
\begin{aligned}
& \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=1-\frac{\mathrm{d} y}{\mathrm{~d} x} \quad \text { A1 } \\
& \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=1(>0) \quad \text { A1 }
\end{aligned}
$$

Note: accept correct reasoning $\frac{\mathrm{d} y}{\mathrm{~d} x}$ that is increasing as $x$ increases.

THEN
hence $(1,1)$ is a local minimum $\quad A G$
[3 marks]

b. integrating factor $=\mathrm{e}^{\int \mathrm{d} x}$
(M1)
$$
\begin{aligned}
& =\mathrm{e}^x \quad \text { (A1) } \\
& \frac{\mathrm{d} y}{\mathrm{~d} x} \mathrm{e}^x+y \mathrm{e}^x=x \mathrm{e}^x \\
& y \mathrm{e}^x=\int x \mathrm{e}^x \mathrm{~d} x \\
& =x \mathrm{e}^x-\int \mathrm{e}^x \mathrm{~d} x \\
& =x \mathrm{e}^x-\mathrm{e}^x(+c)
\end{aligned}
$$
(M1)
Note: Award $\mathbf{A 1}$ for the correct RHS.
substituting $(1,1)$ gives
$$
\begin{aligned}
& \mathrm{e}=\mathrm{e}-\mathrm{e}+c \quad \text { M1 } \\
& c=\mathrm{e} \\
& y=x-1+\mathrm{e}^{1-x}
\end{aligned}
$$
[8 marks]
c.i. METHOD 1
EITHER
attempt to solve for the intersection $x-1+\mathrm{e}^{1-x}=x-1$
(M1)
OR
attempt to find the difference $x-1+\mathrm{e}^{1-x}-(x-1)$
(M1)
THEN
$\mathrm{e}^{1-x}>0$ for all $x \quad \boldsymbol{R 1}$

Note: Accept $\mathrm{e}^{1-x} \neq 0$ or equivalent reasoning.
therefore the curve does not intersect the isocline
$A G$

METHOD 2
$y=x-1$ is an (oblique) asymptote to the curve
R1
Note: Do not accept “the curve is parallel to $y=x-1$ “
$y=x-1$ is the isocline for $\frac{\mathrm{d} y}{\mathrm{~d} x}=1 \quad \boldsymbol{R} \boldsymbol{1}$ therefore the curve does not intersect the isocline
AG
METHOD 3
The initial point is above $y=x-1$, so $\frac{\mathrm{d} y}{\mathrm{~d} x}<1 \quad \boldsymbol{R 1}$
$$
\begin{aligned}
& \Rightarrow x-y<1 \\
& \Rightarrow y>x-1 \quad \text { R1 }
\end{aligned}
$$
therefore the curve does not intersect the isocline
AG
[2 marks]

c.ii.

concave up curve with minimum at approximately $(1,1) \quad A 1$ asymptote of curve is isocline $y=x-1 \quad$ A1

Note: Only award $F \boldsymbol{T}$ from (b) if the above conditions are satisfied.
[2 marks]

 

Question 

Consider the differential equation $2 x y \frac{d y}{d x}=y^2-x^2$, where $x>0$.
a. Solve the differential equation and show that a general solution is $x^2+y^2=c x$ where $c$ is a positive constant.
b. Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$.

▶️Answer/Explanation

Markscheme
a. * This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{y^2-x^2}{2 x y} \\
& \text { let } y=v x \quad \text { M1 } \\
& \frac{d y}{d x}=v+x \frac{d v}{d x} \quad \text { (A1) } \\
& v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 v x^2} \quad \text { (M1) } \\
& v+x \frac{d v}{d x}=\frac{v^2-1}{2 v} \quad\left(=\frac{v}{2}-\frac{1}{2 v}\right)
\end{aligned}
$$
(M1)
Note: Or equivalent attempt at simplification.
$$
\begin{aligned}
& x \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{-v^2-1}{2 v} \quad\left(=-\frac{v}{2}-\frac{1}{2 v}\right) \quad \text { A1 } \\
& \frac{2 v}{1+v^2} \frac{\mathrm{d} v}{\mathrm{~d} x}=-\frac{1}{x} \quad \text { (M1) } \\
& \int \frac{2 v}{1+v^2} \mathrm{~d} v=\int-\frac{1}{x} \mathrm{~d} x \quad \text { (A1) } \\
& \ln \left(1+v^2\right)=-\ln x+\ln c \quad \text { A1A1 }
\end{aligned}
$$
A1
A1A1
Note: Award $\boldsymbol{A} \boldsymbol{1}$ for LHS and $\boldsymbol{A} \boldsymbol{1}$ for RHS and a constant.

$$
\ln \left(1+\left(\frac{y}{x}\right)^2\right)=-\ln x+\ln c \quad M 1
$$
Note: Award $\boldsymbol{M} 1$ for substituting $v=\frac{y}{x}$. May be seen at a later stage.
$$
1+\left(\frac{y}{x}\right)^2=\frac{c}{x} \quad \text { A1 }
$$
Note: Award $\boldsymbol{A} 1$ for any correct equivalent equation without logarithms.
$$
x^2+y^2=c x \quad \boldsymbol{A G}
$$

b. METHOD 1
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y^2-x^2}{2 x y}
$$
(for horizontal tangents) $\frac{d y}{d x}=0 \quad \boldsymbol{M 1}$
$$
\left(\Rightarrow y^2=x^2\right) \Rightarrow y= \pm x
$$
EITHER
using $x^2+y^2=c x \Rightarrow 2 x^2=c x \quad$ M1
$2 x^2-c x=0 \Rightarrow x=\frac{c}{2} \quad$ A1
Note: Award M1A1 for $2 y^2= \pm c y$.
OR
using implicit differentiation of $x^2+y^2=c x$
$$
2 x+2 y \frac{d y}{d x}=c \quad \text { M1 }
$$
Note: Accept differentiation of $y=\sqrt{c x-x^2}$.
$$
\frac{d y}{d x}=0 \Rightarrow x=\frac{c}{2} \quad \text { A1 }
$$
THEN
tangents at $y=\frac{c}{2}, y=-\frac{c}{2} \quad$ A1A1
hence there are two tangents $\quad \boldsymbol{A G}$

METHOD 2
$$
\begin{aligned}
& x^2+y^2=c x \\
& \left(x-\frac{c}{2}\right)^2+y^2=\frac{c^2}{4} \quad \text { M1A1 }
\end{aligned}
$$
this is a circle radius $\frac{c}{2}$ centre $\left(\frac{c}{2}, 0\right) \quad \boldsymbol{A} 1$ hence there are two tangents $A G$ tangents at $y=\frac{c}{2}, y=-\frac{c}{2} \quad$ A1A1

 
 

Question 

A simple model to predict the population of the world is set up as follows. At time $t$ years the population of the world is $x$, which can be assumed to be a continuous variable. The rate of increase of $x$ due to births is $0.056 x$ and the rate of decrease of $x$ due to deaths is $0.035 x$.
a. Show that $\frac{\mathrm{d} x}{\mathrm{~d} t}=0.021 x$.
b. Find a prediction for the number of years it will take for the population of the world to double.

▶️Answer/Explanation

Markscheme
a. * This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
\begin{aligned}
& \frac{\mathrm{d} x}{\mathrm{~d} t}=0.056 x-0.035 x \\
& \frac{\mathrm{d} x}{\mathrm{~d} t}=0.021 x \quad \text { AG }
\end{aligned}
$$
A1
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=0.021 x \quad \text { AG }
$$
[1 mark]
b. METHOD 1
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=0.021 x
$$
attempt to separate variables $\boldsymbol{M 1}$
$$
\begin{aligned}
& \int \frac{1}{x} \mathrm{~d} x=\int 0.021 \mathrm{~d} t \quad \boldsymbol{A 1} \\
& \ln x=0.021 t(+c) \quad \text { A1 }
\end{aligned}
$$
EITHER
$$
\begin{aligned}
& x=A \mathrm{e}^{0.021 t} \\
& \Rightarrow 2 A=A \mathrm{e}^{0.021 t} \quad \boldsymbol{A 1}
\end{aligned}
$$
A1
Note: This $\boldsymbol{A} 1$ is independent of the following marks.
OR

$$
\begin{gathered}
t=0, x=x_0 \Rightarrow c=\ln x_0 \\
\Rightarrow \ln 2 x_0=0.021 t+\ln x_0
\end{gathered}
$$
A1
Note: This $A 1$ is independent of the following marks.
THEN
$$
\begin{aligned}
& \Rightarrow \ln 2=0.021 t \\
& \Rightarrow t=33 \text { years }
\end{aligned}
$$
(M1)
A1
Note: If a candidate writes $t=33.007$, so $t=34$ then award the final $\boldsymbol{A} 1$.
METHOD 2
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=0.021 x
$$
attempt to separate variables $M 1$
$$
\int_A^{2 A} \frac{1}{x} \mathrm{~d} x=\int_0^t 0.021 \mathrm{~d} u
$$
A1A1
Note: Award $\boldsymbol{A} 1$ for correct integrals and $\boldsymbol{A} 1$ for correct limits seen anywhere. Do not penalize use of $t$ in place of $u$.
$$
\begin{aligned}
& {[\ln x]_A^{2 A}=[0.021 u]_0^t \quad \boldsymbol{A 1}} \\
& \Rightarrow \ln 2=0.021 t \quad \text { (M1) } \\
& \Rightarrow t=33 \quad \text { A1 }
\end{aligned}
$$

 
 

Question 

Consider the differential equation $\frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=x^2$, given that $y=2$ when $x=0$.
a. Show that $1+x^2$ is an integrating factor for this differential equation.
b. Hence solve this differential equation. Give the answer in the form $y=f(x)$.

▶️Answer/Explanation

Markscneme
a. ${ }^{\star}$ This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
attempting to find an integrating factor
(M1)
$$
\begin{aligned}
& \int \frac{2 x}{1+x^2} \mathrm{~d} x=\ln \left(1+x^2\right) \quad \text { (M1)A1 } \\
& \mathrm{IF} \text { is } \mathrm{e}^{\ln \left(1+x^2\right)} \quad \text { (M1)A1 } \\
& =1+x^2 \quad \text { AG }
\end{aligned}
$$
(M1)A1
METHOD 2
multiply by the integrating factor
$$
\left(1+x^2\right) \frac{d y}{d x}+2 x y=x^2\left(1+x^2\right) \quad \text { M1A1 }
$$
left hand side is equal to the derivative of $\left(1+x^2\right) y$
A3

$$
\begin{aligned}
& \text { b. }\left(1+x^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y=\left(1+x^2\right) x^2 \quad \text { (M1) } \\
& \frac{\mathrm{d}}{\mathrm{d} x}\left[\left(1+x^2\right) y\right]=x^2+x^4 \\
& \left(1+x^2\right) y=\left(\int x^2+x^4 \mathrm{~d} x=\right) \frac{x^3}{3}+\frac{x^5}{5}(+c) \quad \text { A1A1 } \\
& \left.y=\frac{1}{1+x^2}\left(\frac{x^3}{3}+\frac{x^5}{5}+c\right)\right) \\
& x=0, y=2 \Rightarrow c=2 \quad \text { M1A1 } \\
& y=\frac{1}{1+x^2}\left(\frac{x^3}{3}+\frac{x^5}{5}+2\right) \text { A1 }
\end{aligned}
$$
(M1)

 
 

Question 

Consider the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{y}{x}$, where $x \neq 0$.
Consider the family of curves which satisfy the differential equation $\frac{d y}{d x}=1+\frac{y}{x}$, where $x \neq 0$.
a. Given that $y(1)=1$, use Euler’s method with step length $h=0.25$ to find an approximation for $y(2)$. Give your answer to two significant figures.
b. Solve the equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{y}{x}$ for $y(1)=1$.
c. Find the percentage error when $y(2)$ is approximated by the final rounded value found in part (a). Give your answer to two significant figures.
d.i. Find the equation of the isocline corresponding to $\frac{d y}{d x}=k$, where $k \neq 0, k \in \mathrm{R}$.
d.ii.Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

▶️Answer/Explanation

Markscheme
a. ${ }^*$ This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to apply Euler’s method
(M1)
$$
x_{n+1}=x_n+0.25 ; y_{n+1}=y_n+0.25 \times\left(1+\frac{y_n}{x_n}\right)
$$

Note: Award $\boldsymbol{A} \mathbf{1}$ for correct $x$ values, $\mathbf{A} 1$ for first three correct $y$ values.
$y=3.3 \quad$ A1
[4 marks]
b. METHOD 1
$$
\begin{aligned}
& I(x)=\mathrm{e}^{\int-\frac{1}{x} \mathrm{~d} x} \quad \text { (M1) } \\
& =\mathrm{e}^{-\ln x} \\
& =\frac{1}{x} \quad \text { (A1) } \\
& \frac{1}{x} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{y}{x^2}=\frac{1}{x} \quad \text { (M1) } \\
& \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{y}{x}\right)=\frac{1}{x} \\
& \frac{y}{x}=\ln |x|+C \quad \text { A1 } \\
& y(1)=1 \Rightarrow C=1 \quad \text { M1 } \\
& y=x \ln |x|+x \quad \text { A1 }
\end{aligned}
$$
(M1)
(M1)
A1
METHOD 2
$$
\begin{aligned}
& v=\frac{y}{x} \quad \text { M1 } \\
& \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{1}{x} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{1}{x^2} y \quad \text { (A1) } \\
& v+x \frac{\mathrm{d} v}{\mathrm{~d} x}=1+v \quad \text { M1 } \\
& \int 1 \mathrm{~d} v=\int \frac{1}{x} \mathrm{~d} x \\
& v=\ln |x|+C \\
& \frac{y}{x}=\ln |x|+C \quad \text { A1 } \\
& y(1)=1 \Rightarrow C=1 \quad \text { M1 } \\
& y=x \ln |x|+x \quad \text { A1 }
\end{aligned}
$$
(A1)
[6 marks]
c. $y(2)=2 \ln 2+2=3.38629 \ldots$
$$
\begin{aligned}
& \text { percentage error }=\frac{3.38629 \ldots-3.3}{3.38629 \ldots} \times 100 \% \quad \text { (M1)(A1) } \\
& =2.5 \% \quad \boldsymbol{A 1}
\end{aligned}
$$

d.i. $\frac{\mathrm{d} y}{\mathrm{~d} x}=k \Rightarrow 1+\frac{y}{x}=k \quad$ A1
$$
y=(k-1) x
$$
[1 mark]
d.iigradient of isocline equals gradient of normal
(M1)
$$
\begin{aligned}
& k-1=-\frac{1}{k} \text { or } k(k-1)=-1 \quad \text { A1 } \\
& k^2-k+1=0 \quad \text { A1 } \\
& \Delta=1-4<0 \quad \text { R1 } \\
& \therefore \text { no solution } \quad \text { AG }
\end{aligned}
$$
Note: Accept alternative reasons for no solutions.

 
 

Question 

a. Consider the differential equation
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=f\left(\frac{y}{x}\right), x>0 .
$$
Use the substitution $y=v x$ to show that the general solution of this differential equation is
$$
\int \frac{d v}{f(v)-v}=\ln x+\text { Constant. }
$$
b. Hence, or otherwise, solve the differential equation
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x^2+3 x y+y^2}{x^2}, x>0,
$$
given that $y=1$ when $x=1$. Give your answer in the form $y=g(x)$.

▶️Answer/Explanation

Markscheme
a. ${ }^*$ This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d}{d x} \quad M 1
$$
the differential equation becomes
$$
\begin{aligned}
& v+x \frac{\mathrm{d} v}{\mathrm{~d} x}=f(v) \quad \text { A1 } \\
& \int \frac{\mathrm{d} v}{f(v)-v}=\int \frac{\mathrm{d} v}{x} \quad \text { A1 }
\end{aligned}
$$
integrating, Constant $\int \frac{d v}{f(v)-v}=\ln x+$ Constant $\quad \boldsymbol{A G}$

b. EITHER
$$
\begin{aligned}
& f(v)=1+3 v+v^2 \quad \text { (A1) } \\
& \left(\int \frac{\mathrm{d} v}{f(v)-v}=\right) \int \frac{\mathrm{d} v}{1+3 v+v^2-v}=\ln x+C \quad \text { M1A1 } \\
& \int \frac{\mathrm{d} v}{(1+v)^2}=(\ln x+C) \quad \text { A1 }
\end{aligned}
$$
Note: $\boldsymbol{A 1}$ is for correct factorization.
$$
-\frac{1}{1+v}(=\ln x+C) \quad \boldsymbol{A 1}
$$
OR
$$
v+x \frac{d v}{d x}=1+3 v+v^2
$$
A1
$\int \frac{\mathrm{d} v}{1+2 v+v^2}=\int \frac{1}{x} \mathrm{~d} x \quad$ M1
$$
\int \frac{\mathrm{d} v}{(1+v)^2}\left(=\int \frac{1}{x} \mathrm{~d} x\right)
$$
Note: $\boldsymbol{A 1}$ is for correct factorization.
$$
-\frac{1}{1+v}=\ln x(+C) \quad \text { A1A1 }
$$
substitute $y=1$ or $v=1$ when $x=1$
(M1)
therefore $C=-\frac{1}{2}$
A1
Note: This $\boldsymbol{A 1}$ can be awarded anywhere in their solution.
substituting for $v$,
$-\frac{1}{\left(1+\frac{y}{x}\right)}=\ln x-\frac{1}{2} \quad M 1$
Note: Award for correct substitution of $\frac{y}{x}$ into their expression.
$$
1+\frac{y}{x}=\frac{1}{\frac{1}{2}-\ln x}
$$
Note: Award for any rearrangement of a correct expression that has $y$ in the numerator.

\begin{aligned}
& y=x\left(\frac{1}{\left(\frac{1}{2}-\ln x\right)}-1\right) \text { (or equivalent) } \\
& \left(=x\left(\frac{1+2 \ln x}{1-2 \ln x}\right)\right)
\end{aligned}

 

 
 

Question 

Consider the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x}{x^2+1} y=x$ where $y=1$ when $x=0$.
a. Show that $\sqrt{x^2+1}$ is an integrating factor for this differential equation.
b. Solve the differential equation giving your answer in the form $y=f(x)$.

▶️Answer/Explanation

Markscheme
a. ${ }^*$ This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
integrating factor $=\mathrm{e}^{\int \frac{x}{x^2+1} d x}$
(M1)
$\int \frac{x}{x^2+1} \mathrm{~d} x=\frac{1}{2} \ln \left(x^2+1\right)$
(M1)
Note: Award $\boldsymbol{M} 1$ for use of $u=x^2+1$ for example or $\int \frac{f^{\prime}(x)}{f(x)} \mathrm{d} x=\ln f(x)$.
integrating factor $=\mathrm{e}^{\frac{1}{2} \ln \left(x^2+1\right)} \quad$ A1
$\left.=\mathrm{e}^{\ln \left(\sqrt{x^2+1}\right.}\right) \quad \boldsymbol{A 1}$
Note: Award $\boldsymbol{A} 1$ for $\mathrm{e}^{\ln \sqrt{u}}$ where $u=x^2+1$.
$=\sqrt{x^2+1} \quad A G$

METHOD 2
$$
\begin{aligned}
& \frac{d}{d x}\left(y \sqrt{x^2+1}\right)=\frac{d y}{d x} \sqrt{x^2+1}+\frac{x}{\sqrt{x^2+1}} y \quad \text { M1A1 } \\
& \sqrt{x^2+1}\left(\frac{d y}{d x}+\frac{x}{x^2+1} y\right) \quad \text { M1A1 }
\end{aligned}
$$
Note: Award $\boldsymbol{M} 1$ for attempting to express in the form $\sqrt{x^2+1} \times$ (LHS of de).
so $\sqrt{x^2+1}$ is an integrating factor for this differential equation $\boldsymbol{A G}$
[4 marks]

b. $\sqrt{x^2+1} \frac{d y}{d x}+\frac{x}{\sqrt{x^2+1}} y=x \sqrt{x^2+1}$ (or equivalent)
(M1)
$$
\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}\left(y \sqrt{x^2+1}\right)=x \sqrt{x^2+1} \\
& y \sqrt{x^2+1}=\int x \sqrt{x^2+1} \mathrm{~d} x\left(y=\frac{1}{\sqrt{x^2+1}} \int x \sqrt{x^2+1} \mathrm{~d} x\right) \\
& =\frac{1}{3}\left(x^2+1\right)^{\frac{3}{2}}+C \quad \text { (M1)A1 }
\end{aligned}
$$
A1
Note: Award M1 for using an appropriate substitution.
Note: Condone the absence of $C$.
substituting $x=0, y=1 \Rightarrow C=\frac{2}{3} \quad$ M1
Note: Award $\boldsymbol{M} 1$ for attempting to find their value of $C$.
$$
y=\frac{1}{3}\left(x^2+1\right)+\frac{2}{3 \sqrt{x^2+1}}\left(y=\frac{\left(x^2+1\right)^{\frac{3}{2}}+2}{3 \sqrt{x^2+1}}\right) \quad \boldsymbol{A 1}
$$

 

 
 

Question 

Consider the differential equation $x \frac{y}{d x}-y=x^p+1$ where $x \in \mathrm{R}, x \neq 0$ and $p$ is a positive integer, $p>1$.
a. Solve the differential equation given that $y=-1$ when $x=1$. Give your answer in the form $y=f(x)$.
b.i. Show that the $x$-coordinate(s) of the points on the curve $y=f(x)$ where $\frac{d y}{d x}=0$ satisfy the equation $x^{p-1}=\frac{1}{p}$.
b.ii.Deduce the set of values for $p$ such that there are two points on the curve $y=f(x)$ where $\frac{d y}{d x}=0$. Give a reason for your answer.

▶️Answer/Explanation

Markscheme
a. * This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
$$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}=x^{p-1}+\frac{1}{x} \quad \text { (M1) } \\
& \text { integrating factor }=\mathrm{e}^{\int-\frac{1}{x} \mathrm{~d} x} \quad \text { M1 } \\
& =\mathrm{e}^{-\ln x} \quad \text { (A1) } \\
& =\frac{1}{x} \quad \text { A1 } \\
& \frac{1}{x} \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{y}{x^2}=x^{p-2}+\frac{1}{x^2} \quad \text { (M1) } \\
& \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{y}{x}\right)=x^{p-2}+\frac{1}{x^2} \\
& \frac{y}{x}=\frac{1}{p-1} x^{p-1}-\frac{1}{x}+C \quad \text { A1 }
\end{aligned}
$$
$M 1$
(M1)
A1
Note: Condone the absence of $C$.
$$
y=\frac{1}{p-1} x^p+C x-1
$$
substituting $x=1, y=-1 \Rightarrow C=-\frac{1}{p-1} \quad$ M1

Note: Award $\boldsymbol{M} 1$ for attempting to find their value of $C$.
$y=\frac{1}{p-1}\left(x^p-x\right)-1$
A1
[8 marks]
METHOD 2
put $y=v x$ so that $\frac{\mathrm{d} y}{\mathrm{~d} x}=v+x \frac{\mathrm{d} v}{\mathrm{~d} x} \quad$ M1(A1)
substituting, M1
$x\left(v+x \frac{d v}{d x}\right)-v x=x^p+1$
(A1)
$x \frac{d v}{d x}=x^{p-1}+\frac{1}{x} \quad$ M1
$\frac{\mathrm{d} v}{\mathrm{~d} x}=x^{p-2}+\frac{1}{x^2}$
$v=\frac{1}{p-1} x^{p-1}-\frac{1}{x}+C \quad \boldsymbol{A 1}$
Note: Condone the absence of $C$.
$y=\frac{1}{p-1} x^p+C x-1$
substituting $x=1, y=-1 \Rightarrow C=-\frac{1}{p-1} \quad$ M1
Note: Award $\boldsymbol{M} 1$ for attempting to find their value of $C$.
$y=\frac{1}{p-1}\left(x^p-x\right)-1$
b.i. METHOD 1
find $\frac{d y}{d x}$ and solve $\frac{d y}{d x}=0$ for $x$
$$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{p-1}\left(p x^{p-1}-1\right) \quad \text { M1 } \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \Rightarrow p x^{p-1}-1=0 \quad \boldsymbol{A 1}
\end{aligned}
$$
$$
p x^{p-1}=1
$$
Note: Award a maximum of $M 1 A O$ if a candidate’s answer to part (a) is incorrect.
$$
x^{p-1}=\frac{1}{p} \quad \boldsymbol{A G}
$$
METHOD 2
substitute $\frac{d y}{d x}=0$ and their $y$ into the differential equation and solve for $x$
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=0 \Rightarrow-\left(\frac{x^p-x}{p-1}\right)+1=x^p+1 \quad \text { M1 }
$$
$x^p-x=x^p-p x^p$
A1
$p x^{p-1}=1$
Note: Award a maximum of $M 1 A O$ if a candidate’s answer to part (a) is incorrect.
$$
x^{p-1}=\frac{1}{p} \quad \boldsymbol{A G}
$$

b.ii.there are two solutions for $x$ when $p$ is odd (and $p>1$
A1
if $p-1$ is even there are two solutions (to $x^{p-1}=\frac{1}{p}$ ) and if $p-1$ is odd there is only one solution (to $x^{p-1}=\frac{1}{p}$ ) $\quad \mathbf{R} \mathbf{r}$
Note: Only award the $\boldsymbol{R} 1$ if both cases are considered.

 
 

Question 

This question investigates some applications of differential equations to modeling population growth.
One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{\mathrm{d} P}{\mathrm{~d} t}=k P$, where $k \in \mathrm{R}, t$ is the time (in years) and $P$ is the population
The initial population is 1000 .
Given that $k=0.003$, use your answer from part (a) to find
Consider now the situation when $k$ is not a constant, but a function of time.
Given that $k=0.003+0.002 t$, find
Another model for population growth assumes
– there is a maximum value for the population, $L$.
– that $k$ is not a constant, but is proportional to $\left(1-\frac{P}{L}\right)$.

a. Show that the general solution of this differential equation is $P=A \mathrm{e}^{k t}$, where $A \in \mathrm{R}$.
b.i. the population after 10 years
b.ii.the number of years it will take for the population to triple.
b.iiilim $P$
c.i. the solution of the differential equation, giving your answer in the form $P=f(t)$.
c.ii.the number of years it will take for the population to triple.
d. Show that $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, where $m \in \mathrm{R}$.
e. Solve the differential equation $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, giving your answer in the form $P=g(t)$.
f. Given that the initial population is $1000, L=10000$ and $m=0.003$, find the number of years it will take for the population to triple.

▶️Answer/Explanation

Markscheme
a. $\int \frac{1}{P} \mathrm{~d} P=\int k \mathrm{~d} t \quad$ M1A1
$\ln P=k t+c \quad$ A1A1
$P=e^{k t+c} \quad$ A1
$P=A e^{k t}$, where $A=e^c \quad$ AG
[5 marks]
b.i. when $t=0, P=1000$
$\Rightarrow A=1000 \quad$ A1 $P(10)=1000 e^{0.003(10)}=1030 \quad$ A1
[2 marks]
b.ii. $3000=1000 e^{0.003 t} \quad \boldsymbol{M 1}$
$t=\frac{\ln 3}{0.003}=366$ years $\quad \mathbf{A 1}$
[2 marks]
b.iiilim $P=\infty \quad$ A1
[1 mark]

c.i. $\int \frac{1}{P} \mathrm{~d} P=\int(0.003+0.002 t) \mathrm{d} t \quad M 1$
$$
\ln P=0.003 t+0.001 t^2+c
$$
A1A1
$$
P=e^{0.003 t+0.001 t^2+c}
$$
A1
when $t=0, P=1000$
$$
\Rightarrow e^c=1000 \quad \text { M1 }
$$
$$
P=1000 e^{0.003 t+0.001 t^2}
$$
[5 marks]
c.ii. $3000=1000 e^{0.003 t+0.001 t^2} \quad$ M1
$$
\ln 3=0.003 t+0.001 t^2
$$
A1
Use of quadratic formula or GDC graph or GDC polysmit
$M 1$
$$
t=31.7 \text { years } \quad \text { A1 }
$$
[4 marks]
d. $k=m\left(1-\frac{P}{L}\right)$, where $m$ is the constant of proportionality
A1
So $\frac{\mathrm{d} P}{\mathrm{~d} t}=m\left(1-\frac{P}{L}\right) P \quad$ A1
$$
\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P) \quad \boldsymbol{A G}
$$

e. $\int \frac{1}{P(L-P)} \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t \quad \boldsymbol{M 1}$
$$
\begin{array}{ll}
\frac{1}{P(L-P)}=\frac{A}{P}+\frac{B}{L-P} & \boldsymbol{M 1} \\
1 \equiv A(L-P)+B P & \boldsymbol{A 1}
\end{array}
$$
$$
A=\frac{1}{L}, B=\frac{1}{L} \quad \boldsymbol{A 1}
$$
$$
\frac{1}{L} \int\left(\frac{1}{P}+\frac{1}{L-P}\right) \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t
$$
$\frac{1}{L}(\ln P-\ln (L-P))=\frac{m}{L} t+c \quad$ A1A1
$\ln \left(\frac{P}{L-P}\right)=m t+d$, where $d=c L \quad M 1$
$\frac{P}{L-P}=C e^{m t}$, where $C=e^d \quad \boldsymbol{A 1}$
$$
P\left(1+C e^{m t}\right)=C L e^{m t} \quad \mathbf{M 1}
$$
$P=\frac{C L e^{m t}}{\left(1+C e^{m t}\right)}\left(=\frac{L}{\left(D e^{-m t}+1\right)}\right.$, where $\left.D=\frac{1}{C}\right)$
\begin{aligned}
& \text { f. } 1000=\frac{10000}{D+1} \quad \boldsymbol{M 1} \\
& D=9 \quad \boldsymbol{A 1} \\
& 3000=\frac{10000}{9 e^{-0.003 t}+1} \quad \text { M1 } \\
& t=450 \text { years } \quad \boldsymbol{A 1} \\
&
\end{aligned}

 
 

Question 

Question
The curve $y=f(x)$ has a gradient function given by
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=x-y
$$
The curve passes through the point $(1,1)$.
a.i. On the same set of axes, sketch and label isoclines for $\frac{d y}{d x}=-1,0$ and 1 , and clearly indicate the value of each $y$-intercept.
a.ii.Hence or otherwise, explain why the point $(1,1)$ is a local minimum.
b. Find the solution of the differential equation $\frac{d y}{d x}=x-y$, which passes through the point $(1,1)$. Give your answer in the form $y=f(x)$.
c.i. Explain why the graph of $y=f(x)$ does not intersect the isocline $\frac{d y}{d x}=1$.
c.ii.Sketch the graph of $y=f(x)$ on the same set of axes as part (a)(i).

▶️Answer/Explanation

Markscheme
a.i. attempt to find equation of isoclines by setting $x-y=-1,0,1 \quad \boldsymbol{M 1}$

3 parallel lines with positive gradient
A1
$y$-intercept $=-c$ for $\frac{d y}{d x}=c$
A1
Note: To award $\boldsymbol{A 1}$, each $y$-intercept should be clear, but condone a missing label $(e g .(0,0))$.
If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.
[3 marks]
a.iiat point $(1,1), \frac{d y}{d x}=0 \quad$ A1
EITHER
to the left of $(1,1)$, the gradient is negative
R1
to the right of $(1,1)$, the gradient is positive
R1
Note: Accept any correct reasoning using gradient, isoclines or slope field.
If a candidate uses left/right or $x<1 / x>1$ without explicitly referring to the point $(1,1)$ or a correct region on the diagram, award $\boldsymbol{R O R 1}$.
OR
$$
\begin{aligned}
& \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=1-\frac{\mathrm{d} y}{\mathrm{~d} x} \quad \text { A1 } \\
& \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=1(>0) \quad \text { A1 }
\end{aligned}
$$
A1
A1
Note: accept correct reasoning $\frac{d y}{d x}$ that is increasing as $x$ increases.

THEN
hence $(1,1)$ is a local minimum
AG
[3 marks]
b. integrating factor $=\mathrm{e}^{\int d x}$
(M1)
$$
\begin{aligned}
& =\mathrm{e}^x \quad \text { (A1) } \\
& \frac{\mathrm{d} y}{\mathrm{~d} x} \mathrm{e}^x+y \mathrm{e}^x=x \mathrm{e}^x \quad \text { (M1) } \\
& y \mathrm{e}^x=\int x \mathrm{e}^x \mathrm{~d} x \quad \text { A1 } \\
& =x \mathrm{e}^x-\int \mathrm{e}^x \mathrm{~d} x \quad \text { (M1) } \\
& =x \mathrm{e}^x-\mathrm{e}^x(+c) \quad \text { A1 }
\end{aligned}
$$
A1
A1
Note: Award $\boldsymbol{A} 1$ for the correct RHS.
substituting $(1,1)$ gives
$$
\begin{aligned}
& \mathrm{e}=\mathrm{e}-\mathrm{e}+c \quad \text { M1 } \\
& c=\mathrm{e} \\
& y=x-1+\mathrm{e}^{1-x} \quad \text { A1 }
\end{aligned}
$$
[8 marks]

C.i. METHOD 1
EITHER
attempt to solve for the intersection $x-1+\mathrm{e}^{1-x}=x-1$
(M1)
OR
attempt to find the difference $x-1+\mathrm{e}^{1-x}-(x-1)$
(M1)
THEN
$\mathrm{e}^{1-x}>0$ for all $x \quad R 1$
Note: Accept $\mathrm{e}^{1-x} \neq 0$ or equivalent reasoning.
therefore the curve does not intersect the isocline $\boldsymbol{A G}$
METHOD 2
$y=x-1$ is an (oblique) asymptote to the curve $R 1$
Note: Do not accept “the curve is parallel to $y=x-1$ “
$y=x-1$ is the isocline for $\frac{d y}{d x}=1 \quad \boldsymbol{R} 1$
therefore the curve does not intersect the isocline $\boldsymbol{A G}$

METHOD 3
The initial point is above $y=x-1$, so $\frac{\mathrm{d} y}{\mathrm{~d} x}<1 \quad$ R1
$$
\begin{aligned}
& \Rightarrow x-y<1 \\
& \Rightarrow y>x-1 \quad \text { R1 }
\end{aligned}
$$
therefore the curve does not intersect the isocline $\quad \boldsymbol{A G}$
[2 marks]

c.ii.

concave up curve with minimum at approximately $(1,1)$
A1 asymptote of curve is isocline $y=x-1 \quad \boldsymbol{A 1}$
Note: Only award $\boldsymbol{F T}$ from (b) if the above conditions are satisfied.
[2 marks]

 

 

 
 

Question

Given that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2{y^2} = {{\text{e}}^x}\) and y = 1 when x = 0, use Euler’s method with a step length of 0.1 to find an approximation for the value of y when x = 0.4. Give all intermediate values with maximum possible accuracy.

▶️Answer/Explanation

Markscheme

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x} + 2{y^2}\)     (A1)

 

required approximation = 3.85     A1

[8 marks]

Examiners report

Most candidates seemed familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy despite the advice in the question or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures.

Question

Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2} + {x^2}}}{{2{x^2}}}\) for which y = −1 when x = 1.

(a)     Use Euler’s method with a step length of 0.25 to find an estimate for the value of y when x = 2 .

(b)     (i)     Solve the differential equation giving your answer in the form \(y = f(x)\) .

         (ii)     Find the value of y when x = 2 .

▶️Answer/Explanation

Markscheme

(a)     Using an increment of 0.25 in the x-values     A1

Note: The A1 marks are awarded for final column.

\( \Rightarrow y(2) \approx – 0.304\)     A1

[7 marks]

 

(b)     (i)     let y = vx     M1

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (A1)

\( \Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} + {x^2}}}{{2{x^2}}}\)     (M1)

\( \Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{1 – 2v + {v^2}}}{2}\)     (A1)

\( \Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{{(1 – v)}^2}}}{2}\)     A1

\( \Rightarrow \int {\frac{2}{{{{(1 – v)}^2}}}{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} } \)     M1

\( \Rightarrow 2{(1 – v)^{ – 1}} = \ln x + c\)     A1A1

\( \Rightarrow \frac{2}{{1 – \frac{y}{x}}} = \ln x + c\)

when \(x = 1,{\text{ }}y = – 1 \Rightarrow c = 1\)     M1A1

\( \Rightarrow \frac{{2x}}{{x – y}} = \ln x + 1\)

\( \Rightarrow y = x – \frac{{2x}}{{1 + \ln x}}{\text{ }}\left( { = \frac{{x\ln x – x}}{{1 + \ln x}}} \right)\)     M1A1

 

(ii)     when \(x = 2,{\text{ }}y = – 0.362\,\,\,\,\,\left( {{\text{accept 2}} – \frac{4}{{1 + \ln 2}}} \right)\)     A1

[13 marks]

Total [20 marks]

Examiners report

Part (a) was well done by many candidates, but a number were penalised for not using a sufficient number of significant figures. Part (b) was started by the majority of candidates, but only the better candidates were able to reach the end. Many were unable to complete the question correctly because they did not know what to do with the substitution y = vx and because of arithmetic errors and algebraic errors.

Question

Given that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2{y^2} = {{\text{e}}^x}\) and y = 1 when x = 0, use Euler’s method with a step length of 0.1 to find an approximation for the value of y when x = 0.4. Give all intermediate values with maximum possible accuracy.

▶️Answer/Explanation

Markscheme

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x} + 2{y^2}\)     (A1)

 

 

required approximation = 3.85     A1

[8 marks]

Examiners report

Most candidates seemed familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy despite the advice in the question or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures.

Question

The real and imaginary parts of a complex number \(x + {\text{i}}y\) are related by the differential equation \((x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x – y) = 0\).

By solving the differential equation, given that \(y = \sqrt 3 \) when x =1, show that the relationship between the modulus r and the argument \(\theta \) of the complex number is \(r = 2{{\text{e}}^{\frac{\pi }{3} – \theta }}\).

▶️Answer/Explanation

Markscheme

\((x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x – y) = 0\)

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y – x}}{{x + y}}\)

let \(y = vx\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     A1

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx – x}}{{x + vx}}\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{v – 1}}{{v + 1}} – v = \frac{{v – 1 – {v^2} – v}}{{v + 1}} = \frac{{ – 1 – {v^2}}}{{1 + v}}\)     A1

\(\int {\frac{{v + 1}}{{1 + {v^2}}}{\text{d}}v = – \int {\frac{1}{x}{\text{d}}x} } \)     M1

\(\int {\frac{v}{{1 + {v^2}}}{\text{d}}v + \int {\frac{1}{{1 + {v^2}}}{\text{d}}v = – \int {\frac{1}{x}{\text{d}}x} } } \)     M1

\( \Rightarrow \frac{1}{2}\ln \left| {1 + {v^2}} \right| + \arctan v = – \ln \left| x \right| + k\)     A1A1

Notes: Award A1 for \(\frac{1}{2}\ln \left| {1 + {v^2}} \right|\), A1 for the other two terms.

Do not penalize missing k or missing modulus signs at this stage.

 

\( \Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = – \ln \left| x \right| + k\)     M1

\( \Rightarrow \frac{1}{2}\ln 4 + \arctan \sqrt 3 = – \ln 1 + k\)     (M1)

\( \Rightarrow k = \ln 2 + \frac{\pi }{3}\)     A1

\( \Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = – \ln \left| x \right| + \ln 2 + \frac{\pi }{3}\)

attempt to combine logarithms     M1

\( \Rightarrow \frac{1}{2}\ln \left| {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right| + \frac{1}{2}\ln \left| {{x^2}} \right| = \ln 2 + \frac{\pi }{3} – \arctan \frac{y}{x}\)

\( \Rightarrow \frac{1}{2}\ln \left| {{y^2} + {x^2}} \right| = \ln 2 + \frac{\pi }{3} – \arctan \frac{y}{x}\)     (A1)

\( \Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2 + \frac{\pi }{3}\arctan \frac{y}{x}}}\)     (A1)

\( \Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2}} \times {{\text{e}}^{\frac{\pi }{3} – \arctan \frac{y}{x}}}\)     A1

\( \Rightarrow r = 2{{\text{e}}^{\frac{\pi }{3} – \theta }}\)     AG

[15 marks]

Examiners report

Most candidates realised that this was a homogeneous differential equation and that the substitution \(y = vx\) was the way forward. Many of these candidates reached as far as separating the variables correctly but integrating \(\frac{{v + 1}}{{{v^2} + 1}}\) proved to be too difficult for many candidates – most failed to realise that the expression had to be split into two separate integrals. Some candidates successfully evaluated the arbitrary constant but the combination of logs and the subsequent algebra necessary to obtain the final result proved to be beyond the majority of candidates.

Question

Solve the differential equation

\[{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} + xy + 4{x^2},\]

given that y = 2 when x =1. Give your answer in the form \(y = f(x)\).

▶️Answer/Explanation

Markscheme

put \(y = vx\) so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (M1)

the equation becomes \(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2} + v + 4\)     A1

\(\int {\frac{{{\text{d}}v}}{{{v^2} + 4}} = \int {\frac{{{\text{d}}x}}{x}} } \)     A1

\(\frac{1}{2}\arctan \left( {\frac{v}{2}} \right) = \ln x + C\)     A1A1

substituting\((x,{\text{ }}v) = (1,{\text{ }}2)\)

\(C = \frac{\pi }{8}\)     M1A1

the solution is

\(\arctan \left( {\frac{y}{{2x}}} \right) = 2\ln x + \frac{\pi }{4}\)     A1

\(y = 2x\tan \left( {2\ln x + \frac{\pi }{4}} \right)\)     A1

[9 marks]

Examiners report

Most candidates recognised this differential equation as one in which the substitution \(y = vx\) would be helpful and many carried the method through to a successful conclusion. The most common error seen was an incorrect integration of \(\frac{1}{{4 + {v^2}}}\) with partial fractions and/or a logarithmic evaluation seen. Some candidates failed to include an arbitrary constant which led to a loss of marks later on.

Question

Consider the differential equation

\[x\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2y = \frac{{{x^3}}}{{{x^2} + 1}}.\]

(a)     Find an integrating factor for this differential equation.

(b)     Solve the differential equation given that \(y = 1\) when \(x = 1\) , giving your answer in the forms \(y = f(x)\) .

▶️Answer/Explanation

Markscheme

(a)     Rewrite the equation in the form

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{2}{x}y = \frac{{{x^2}}}{{{x^2} + 1}}\)     M1A1

Integrating factor \( = {{\text{e}}^{\int { – \frac{2}{x}{\text{d}}x} }}\)     M1

\( = {{\text{e}}^{ – 2\ln x}}\)     A1

\( = \frac{1}{{{x^2}}}\)     A1

Note: Accept \(\frac{1}{{{x^3}}}\) as applied to the original equation.

 

[5 marks]

 

(b)     Multiplying the equation,

\(\frac{1}{{{x^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{2}{{{x^3}}}y = \frac{1}{{{x^2} + 1}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{{{x^2}}}} \right) = \frac{1}{{{x^2} + 1}}\)     (M1)(A1)

\(\frac{y}{{{x^2}}} = \int {\frac{{{\text{d}}x}}{{{x^2} + 1}}} \)     M1

\( = \arctan x + C\)     A1

Substitute \(x = 1,{\text{ }}y = 1\) .     M1

\(1 = \frac{\pi }{4} + C \Rightarrow C = 1 – \frac{\pi }{4}\)     A1

\(y = {x^2}\left( {\arctan x + 1 – \frac{\pi }{4}} \right)\)     A1

[8 marks]

Total [13 marks]

Examiners report

The response to this question was often disappointing. Many candidates were unable to find the integrating factor successfully. 

Question (2021)

In this question you will be exploring the strategies required to solve a system of linear differential equations.
Consider the system of linear differential equations of the form:
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=x-y \text { and } \frac{\mathrm{d} y}{\mathrm{~d} t}=a x+y
$$
where $x, y, t \in \mathbb{R}^{+}$and $a$ is a parameter.
First consider the case where $a=0$.
Now consider the case where $a=-1$.
Now consider the case where $a=-4$.
From previous cases, we might conjecture that a solution to this differential equation is $y=F \mathrm{e}^{\lambda t}, \lambda \in \mathbb{R}$ and $F$ is a constant.
a.i. By solving the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} t}=y$, show that $y=A \mathrm{e}^t$ where $A$ is a constant.
a.ii.Show that $\frac{\mathrm{d} x}{\mathrm{~d} t}-x=-A \mathrm{e}^t$
a.iiiSolve the differential equation in part (a)(ii) to find $x$ as a function of $t$.
b.i. By differentiating $\frac{\mathrm{d} y}{\mathrm{~d} t}=-x+y$ with respect to $t$, show that $\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=2 \frac{\mathrm{d} y}{\mathrm{~d} t}$.

b.ii.By substituting $Y=\frac{d y}{d t}$, show that $Y=B \mathrm{e}^{2 t}$ where $B$ is a constant.
b.iiit Hence find $y$ as a function of $t$.
b.ivHence show that $x=-\frac{B}{2} \mathrm{e}^{2 t}+C$, where $C$ is a constant.
c.i. Show that $\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=0$.
c.ii.Find the two values for $\lambda$ that satisfy $\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=0$.
c.iiiLet the two values found in part (c)(ii) be $\lambda_1$ and $\lambda_2$.
Verify that $y=\mathrm{Fe}^{\lambda_1 t}+\mathrm{Ge}^{\lambda_2 t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

▶️Answer/Explanation

Markscheme
a.i. METHOD 1
$$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{dt}}=y \\
& \int \frac{\mathrm{d} y}{y}=\int \mathrm{d} \mathrm{t} \quad \text { (M1) } \\
& \ln y=t+c \text { OR } \ln |y|=t+c
\end{aligned}
$$
(M1)
A1A1
Note: Award $\boldsymbol{A} 1$ for $\ln y$ and $\boldsymbol{A} 1$ for $t$ and $c$.
$$
y=A \mathrm{e}^t \quad A G
$$
METHOD 2
rearranging to $\frac{\mathrm{d} y}{\mathrm{dt}}-y=0$ AND multiplying by integrating factor $\mathrm{e}^{-t} \quad \boldsymbol{M} 1$
$$
\begin{array}{ll}
y \mathrm{e}^{-t}=A & \text { A1A1 } \\
y=A \mathrm{e}^t & \text { AG }
\end{array}
$$
[3 marks]

a.iisubstituting $y=A \mathrm{e}^t$ into differential equation in $x$
M1
$$
\begin{aligned}
& \frac{\mathrm{d} x}{\mathrm{dt}}=x-A \mathrm{e}^t \\
& \frac{\mathrm{d} x}{\mathrm{dt}}-x=-A \mathrm{e}^t \quad \text { AG }
\end{aligned}
$$
[1 mark]
a.iiiintegrating factor $(\mathrm{IF})$ is $\mathrm{e}^{\int-1 \mathrm{~d} t}$
$$
\begin{aligned}
& =\mathrm{e}^{-t} \\
& \mathrm{e}^{-t} \frac{\mathrm{d} x}{\mathrm{dt}}-x \mathrm{e}^{-t}=-A \\
& x \mathrm{e}^{-t}=-A t+D \\
& x=(-A t+D) \mathrm{e}^t
\end{aligned}
$$
(A1)
(M1)
(A1)
A1
Note: The first constant must be $A$, and the second can be any constant for the final $\boldsymbol{A} \boldsymbol{1}$ to be awarded. Accept a change of constant applied at the end.
[4 marks]

b.i. $\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=-\frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{\mathrm{d} y}{\mathrm{~d} t}$
A1
EITHER
$$
\begin{aligned}
& =-x+y+\frac{\mathrm{d} y}{\mathrm{~d} t} \\
& =\frac{\mathrm{d} y}{\mathrm{~d} t}+\frac{\mathrm{d} y}{\mathrm{~d} t} \quad \text { (M1) }
\end{aligned}
$$
A1
OR
$$
\begin{aligned}
& =-x+y+(-x+y) \\
& =2(-x+y) \quad \text { (M1) }
\end{aligned}
$$
(M1)
A1
THEN
$$
=2 \frac{\mathrm{d} y}{\mathrm{~d} t} \quad \quad \boldsymbol{A G}
$$
[3 marks]
b.ii. $\frac{\mathrm{d} Y}{\mathrm{~d} t}=2 Y$
A1
$\int \frac{d Y}{Y}=\int 2 d t$
$M 1$
$\ln |Y|=2 t+c$ OR $\ln Y=2 t+c$
A1
$$
Y=B \mathrm{e}^{2 t} \quad \quad \boldsymbol{A G}
$$
[3 marks]

$$
\begin{aligned}
& \text { b.iii } \frac{\mathrm{d} y}{\mathrm{~d} t}=B \mathrm{e}^{2 t} \\
& y=\int B \mathrm{e}^{2 t} \mathrm{~d} t \quad \quad M 1 \\
& y=\frac{B}{2} \mathrm{e}^{2 t}+C \quad \text { A1 } \\
&
\end{aligned}
$$
Note: The first constant must be $B$, and the second can be any constant for the final $\boldsymbol{A} \boldsymbol{1}$ to be awarded. Accept a change of constant applied at the end.
[2 marks]
b.iVMETHOD 1
$$
\begin{aligned}
& B e^{2 t}=-x+\frac{B}{2} e^{2 t}+C \\
& x=-\frac{B}{2} e^{2 t}+C \quad A G
\end{aligned}
$$
A1
Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the $\boldsymbol{A G}$.

METHOD 2
$$
\begin{aligned}
& \frac{\mathrm{d} x}{\mathrm{~d} t}=x-\frac{B}{2} \mathrm{e}^{2 t}-C \\
& \frac{\mathrm{d} x}{\mathrm{~d} t}-x=-\frac{B}{2} \mathrm{e}^{2 t}-C \\
& \frac{\mathrm{d}\left(x \mathrm{e}^{-t}\right)}{\mathrm{d} t}=-\frac{B}{2} \mathrm{e}^t-C \mathrm{e}^{-t} \\
& x \mathrm{e}^{-t}=\int-\frac{B}{2} \mathrm{e}^t-C \mathrm{e}^{-t} \mathrm{~d} t \\
& x \mathrm{e}^{-t}=-\frac{B}{2} \mathrm{e}^t-C \mathrm{e}^{-t}+D \\
& x=-\frac{B}{2} \mathrm{e}^{2 t}+C+D \mathrm{e}^t \\
& \frac{\mathrm{d} y}{\mathrm{~d} t}=-x+y \Rightarrow B \mathrm{e}^{2 t}=\frac{B}{2} \mathrm{e}^{2 t}-C-D \mathrm{e}^t+\frac{B}{2} \mathrm{e}^{2 t}+C \Rightarrow D=0 \\
& \boldsymbol{x}=-\frac{B}{2} e^{2 t}+C
\end{aligned}
$$
[3 marks]
c.i. $\frac{\mathrm{d} y}{\mathrm{~d} t}=-4 x+y$
$\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=-4 \frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{\mathrm{d} y}{\mathrm{~d} t}$ seen anywhere $\quad$ M1

METHOD 1
$$
\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=-4(x-y)+\frac{\mathrm{d} y}{\mathrm{~d} t}
$$
attempt to eliminate $x \quad M 1$
$$
\begin{aligned}
& =-4\left(\frac{1}{4}\left(y-\frac{\mathrm{d} y}{\mathrm{~d} t}\right)-y\right)+\frac{\mathrm{d} y}{\mathrm{~d} t} \\
& =2 \frac{\mathrm{d} y}{\mathrm{~d} t}+3 y \quad \text { A1 } \\
& \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=0 \quad \text { AG }
\end{aligned}
$$
METHOD 2
rewriting LHS in terms of $x$ and $y \quad M 1$
$$
\begin{aligned}
& \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=(-8 x+5 y)-2(-4 x+y)-3 y \quad \text { A1 } \\
& =0 \quad \text { AG }
\end{aligned}
$$
[3 marks]

$$
\begin{gathered}
\text { c.ii. } \frac{\mathrm{d} y}{\mathrm{~d} t}=F \lambda \mathrm{e}^{\lambda t}, \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=F \lambda^2 \mathrm{e}^{\lambda t} \\
F \lambda^2 \mathrm{e}^{\lambda t}-2 F \lambda \mathrm{e}^{\lambda t}-3 F \mathrm{e}^{\lambda t}=0 \\
\lambda^2-2 \lambda-3=0 \quad\left(\text { since } \mathrm{e}^{\lambda t} \neq 0\right)
\end{gathered}
$$
A1
$\lambda_1$ and $\lambda_2$ are 3 and -1 (either order)
A1
[4 marks]
c.iiiMETHOD 1
$$
\begin{aligned}
& y=F \mathrm{e}^{3 t}+G \mathrm{e}^{-t} \\
& \frac{\mathrm{d} y}{\mathrm{~d} t}=3 F \mathrm{e}^{3 t}-G \mathrm{e}^{-t}, \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=9 \mathrm{e}^{3 t}-G \mathrm{e}^{-t} \quad \text { (A1)(A1) } \\
& \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=9 F \mathrm{e}^{3 t}+G \mathrm{e}^{-t}-2\left(3 F \mathrm{e}^{3 t}-G \mathrm{e}^{-t}\right)-3\left(F \mathrm{e}^{3 t}-G \mathrm{e}^{-t}\right) \\
& =9 F \mathrm{e}^{3 t}+G \mathrm{e}^{-t}-6 F \mathrm{e}^{3 t}+2 G \mathrm{e}^{-t}-3 F \mathrm{e}^{3 t}-3 G \mathrm{e}^{-t} \\
& =0 \quad \text { AG }
\end{aligned}
$$

METHOD 2
$$
\begin{aligned}
& y=F \mathrm{e}^{\lambda_1 t}+G \mathrm{e}^{\lambda_2 t} \\
& \frac{\mathrm{d} y}{\mathrm{~d} t}=F \lambda_1 \mathrm{e}^{\lambda_1 t}+G \lambda_2 \mathrm{e}^{\lambda_2 t}, \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}=F \lambda_1^2 \mathrm{e}^{\lambda_1 t}+G \lambda_2^2 \mathrm{e}^{\lambda_2 t} \\
& \frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} t}-3 y=F \lambda_1^2 \mathrm{e}^{\lambda_1 t}+G \lambda_2^2 \mathrm{e}^{\lambda_2 t}-2\left(F \lambda_1 \mathrm{e}^{\lambda_1 t}+G \lambda_2 \mathrm{e}^{\lambda_2 t}\right)-3\left(F \mathrm{e}^{\lambda_1 t}+G \mathrm{e}^{\lambda_2 t}\right) \\
& =F \mathrm{e}^{\lambda_1 t}\left(\lambda^2-2 \lambda-3\right)+G \mathrm{e}^{\lambda_2 t}\left(\lambda^2-2 \lambda-3\right) \quad \text { (A1)(A1) } \\
& =0 \quad \text { AG }
\end{aligned}
$$
(A1)(A1)
[4 marks]

 

 

 
 
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