IB Mathematics AHL 5.18 differential equations AA HL Paper 3
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).
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)
$$
\begin{aligned}
& \Rightarrow y^2-x y=0 \\
& y(y-x)=0 \\
& y=0 \text { or } y=x
\end{aligned}
$$
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 }
$$
Note: Award $\boldsymbol{M} 1$ for using initial conditions to find $c$.
A1
[1 mark]
c.ii.recognition of homogeneous equation,
let $y=v x \quad M 1$
the equation can be written as
$$
\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}
$$
\begin{aligned}
& \Rightarrow c=\frac{\pi}{6 \sqrt{3}}(=0.302) \quad \text { A1 } \\
& \arctan \left(\frac{v-1}{\sqrt{3}}\right)=\sqrt{3} \ln x+\frac{\pi}{6} \\
& \text { substituting } v=\frac{y}{x} \quad \text { M1 }
\end{aligned}
$$
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 $\boldsymbol{A} 1$.
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 $\boldsymbol{A} 1$.