Show that $\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)$.

Show that the differential equation $f\left( x \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}$ can be written as $f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z$.

Hence show that the solution to the differential equation $\left( {x - 3} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}$ given that $x = 4$ when $y = 5$ is $\frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}$.

$z = \frac{{y - x}}{{y + x}}$

$\Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{\left( {y + x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} - 1} \right) - \left( {y - x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + 1} \right)}}{{{{\left( {y + x} \right)}^2}}}$      M1A1

$\Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y - x - y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y + x}}{{{{\left( {y + x} \right)}^2}}}$     A1

$\Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)$     AG

[3 marks]

$f\left( x \right)\left( {\frac{{{{\left( {y + x} \right)}^2}}}{2}} \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = {y^2} - {x^2}$      (M1)

$f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{{{\left( {y + x} \right)}^2}}}$      A1

$f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)}}{{\left( {y + x} \right)}} = 2z$      AG

[2 marks]

METHOD 1

$f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z$

$\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{f\left( x \right)}}$

$\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{x - 3}}$     M1A1

EITHER

$\Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + c$     A1

when $y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}$     M1

$\Rightarrow c = {\text{ln}}\frac{1}{9}$      A1

$\Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\frac{1}{9}$

$\Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {x - 3} \right)^2} - {\text{ln}}\,9$     A1

$\Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {\frac{{x - 3}}{3}} \right)^2}$     A1

$\Rightarrow z = {\left( {\frac{{x - 3}}{3}} \right)^2}$

OR

$\Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\,c$    A1

$z = c{\left( {x - 3} \right)^2}$     M1A1

when $y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}$     M1

$\Rightarrow c = \frac{1}{9}$      A1

THEN

$\Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}$      AG

METHOD 2

$f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z$

$f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} - 2z = 0$

$\frac{{{\text{d}}z}}{{{\text{d}}x}} - \frac{{2z}}{{x - 3}} = 0$     M1

integrating factor is ${e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}}$     A1

${e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}} = {e^{ - 2\,{\text{ln}}\left( {x - 3} \right)}}$

$= \frac{1}{{{{\left( {x - 3} \right)}^2}}}$      A1

hence $\frac{{\text{d}}}{{{\text{d}}x}}\left[ {\frac{z}{{{{\left( {x - 3} \right)}^2}}}} \right] = 0$     M1

$z = A{\left( {x - 3} \right)^2}$     A1

when when $y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}$     M1

$\Rightarrow A = \frac{1}{9}$     A1

$\Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}$      AG

[7 marks]