Show that putting $z = {y^2}$ transforms the differential equation into $\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2xz = 2x$.

By solving this differential equation in $z$, obtain an expression for $y$ in terms of $x$.

METHOD 1

$z = {y^2} \Rightarrow y = {z^{1/2}}$

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{2{z^{1/2}}}}\frac{{{\text{d}}z}}{{{\text{d}}x}}$    M1A1

substituting, $\frac{1}{{2{z^{1/2}}}}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{x}{{{z^{1/2}}}} - x{z^{1/2}}$     M1A1

$\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2xz = 2x$    AG

METHOD 2

$z = {y^2}$

$\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}$    M1A1

$\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2x - 2x{y^2}$    M1A1

$\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2xz = 2x$    AG

[4 marks]

METHOD 1

integrating factor $= {{\text{e}}^{\int {2x{\text{d}}x} }} = {{\text{e}}^{{x^2}}}$     (M1)A1

${{\text{e}}^{{x^2}}}\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2x{{\text{e}}^{{x^2}}}z = 2x{{\text{e}}^{{x^2}}}$    (M1)

$z{{\text{e}}^{{x^2}}} = \int {2x{{\text{e}}^{{x^2}}}{\text{d}}x}$    A1

$= {{\text{e}}^{{x^2}}} + C$    A1

substitute $y = 2$ therefore $z = 4$ when $x = 0$     (M1)

$4 = 1 + C$

$C = 3$    (A1)

the solution is $z = 1 + 3{{\text{e}}^{ - {x^2}}}$     (M1)

Note:     This line may be seen before determining the value of $C$.

so that $y = \sqrt {1 + 3{{\text{e}}^{ - {x^2}}}}$     A1

METHOD 2

$\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2x(1 - z)$

$\int {\frac{1}{{1 - z}}{\text{d}}z = \int {2x{\text{d}}x} }$    M1

$- \ln (1 - z) = {x^2} + C$    A1A1

$1 - z = {{\text{e}}^{ - {x^2} - c}}$ (or $1 - z = B{{\text{e}}^{ - {x^2}}}$)     M1A1

solving for $z$     (M1)

$z = 1 + A{{\text{e}}^{ - {x^2}}}$

$z = 4$ when $x = 0$     (M1)

so $A = 3$     (A1)

the solution is $z = 1 + 3{{\text{e}}^{ - {x^2}}}$

so $y = \sqrt {1 + 3{{\text{e}}^{ - {x^2}}}}$     A1

[9 marks]

Several misconceptions were identified that showed poor understanding of the chain rule. Although many candidates were successful in establishing the result the presentation of their work was far from what is expected in a show that question.

Part (b) was well attempted using both method 1 (integration factor) and 2 (separation of variables). The most common error was omission of the constant of integration or errors in finding its value. Candidates that used method 2 often had difficulties in integrating $\frac{1}{{(1 - z)}}$ correctly and making $z$ the subject often losing out on accuracy marks.