Use Euler’s method with step length 0.1 to find an approximate value of $y$ when $x = 0.4$.

Show that $\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}$.

Show that $\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} =  - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)$.

Find the Maclaurin expansion for $y$ up to and including the term in ${{x^3}}$.

Euler’s method with step length $h = 0.1$ to find $y$ when $x = 0.4$

Note: Accept 3 significant figures in the table.

first line of table       (M1)(A1)

line 2        (A1)

line 3        (A1)

hence $y$ = 3.65       A1

Note: Accept any answer that rounds to 3.65.

[5 marks]

$\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 5y$

$\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1A1A1

Note: Award M1 for a valid attempt to differentiate, A1 for LHS, A1 for RHS.

$\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}} - 5\frac{{{\text{d}}y}}{{{\text{d}}x}} - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}$

$\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}$      AG

[3 marks]

$\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}$

$\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} =  - 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)$     M1A1A1A1

$\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} =  - 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right) - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)$

$\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} =  - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)$     AG

[4 marks]

when $x = 0\,\,\,y = 2$

when $x = 0\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = 5$      A1

when $x = 0\,\,\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} =  - 12$      A1

when $x = 0\,\,\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = 120$      A1

Note: Allow follow through from incorrect values of derivatives.

$y = 2 + 5x - 6{x^2} + 20{x^3}$      M1A1

[5 marks]