Consider the differential equation $z^2 \frac{\mathrm{d}w}{\mathrm{d}z} = w^2 - 2z^2$ for $z > 0$ and $w > 2z$.

• Attempt to use Euler's technique M1

• Correct formulation of Euler's method: $z_{n+1} = z_{n} + 0.1$ $w_{n+1} = w_{n} + 0.1 \times \frac{\mathrm{d}w}{\mathrm{d}z}$, where $\frac{\mathrm{d}w}{\mathrm{d}z} = \frac{w^2 - 2z^2}{z^2}$ A1

• Correct intermediate $w$-values: $3.7, 4.63140..., 5.92098..., 7.79542...$ A1

A1 for any two correct $w$-values seen

• Final answer: $w = 10.7$ A1

For the final A1, the value 10.7 must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.

4 marks total

• $w=vz \Rightarrow \frac{\mathrm{d}w}{\mathrm{d}z}=v+z \frac{\mathrm{d}v}{\mathrm{d}z}$ 1 mark

• Replacing $w$ with $vz$ and $\frac{\mathrm{d}w}{\mathrm{d}z}$ with $v+z \frac{\mathrm{d}v}{\mathrm{d}z}$ in the original equation:

  $z^2 \frac{\mathrm{d}w}{\mathrm{d}z}=w^2-2z^2 \Rightarrow z^2\left(v+z \frac{\mathrm{d}v}{\mathrm{d}z}\right)=v^2z^2-2z^2$ 1 mark

• $v+z \frac{\mathrm{d}v}{\mathrm{d}z}=v^2-2 \quad(\text{since } z>0)$

• $z \frac{\mathrm{d}v}{\mathrm{d}z}=v^2-v-2$ 1 mark

3 marks total

• Attempt to separate variables $w$ and $z$: M1

  $\int \frac{\mathrm{d} w}{w^{2}-2z^2}=\int \frac{\mathrm{d} z}{z^2}$

• Substitute $v = \frac{w}{z}$ (seen anywhere): M1

  $\int \frac{\mathrm{d} v}{v^{2}-1-2}=\int \frac{\mathrm{d} z}{z}$

• Express in partial fraction form: M1

  $\frac{1}{(v-2)(v+1)} = \frac{1}{3}\left(\frac{1}{v-2}-\frac{1}{v+1}\right)$

• Integrate both sides: A1

  $\frac{1}{3}(\ln |v-2|-\ln |v+1|)=\ln |z| + c$

Condone absence of modulus signs throughout.

Method #1

• Attempt to find $c$ using $z=1, w=3, v=3$: M1

  $c=\frac{1}{3} \ln \frac{1}{4}$

• Express both sides as a single logarithm: A1

  $\ln \left|\frac{v-2}{v+1}\right|=\ln \left(\frac{|z|^{3}}{4}\right)$

Method #2

• Express both sides as a single logarithm:

  $\ln \left|\frac{v-2}{v+1}\right|=\ln \left(A|z|^{3}\right)$

• Attempt to find $A$ using $z=1, w=3, v=3$:

  $A=\frac{1}{4}$

Continuation

• Simplify: A1

  $\left|\frac{v-2}{v+1}\right|=\frac{1}{4} z^{3}$ (since $z>0$)

• Substitute back $v=\frac{w}{z}$:

  $\frac{\frac{w}{z}-2}{\frac{w}{z}+1}=\frac{1}{4} z^{3}$ (since $w>2z$)

• Attempt to make $w$ the subject: M1

  $w-\frac{z^{3} w}{4}=2 z+\frac{z^{4}}{4}$

• Final answer: A1

  $w=\frac{8 z+z^{4}}{4-z^{3}}$

8 marks total