Show that $\sqrt {{x^2} + 1}$ is an integrating factor for this differential equation.

Solve the differential equation giving your answer in the form $y = f(x)$.

METHOD 1

integrating factor $= {{\text{e}}^{\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} }}$     (M1)

$\int {\frac{x}{{{x^2} + 1}}{\text{d}}x = \frac{1}{2}\ln ({x^2} + 1)}$     (M1)

Note:     Award M1 for use of $u = {x^2} + 1$ for example or $\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \ln f(x)}$.

integrating factor $= {{\text{e}}^{\frac{1}{2}\ln ({x^2} + 1)}}$     A1

$= {{\text{e}}^{\ln \left( {\sqrt {{x^2} + 1} } \right)}}$     A1

Note:     Award A1 for ${{\text{e}}^{\ln \sqrt u }}$ where $u = {x^2} + 1$.

$= \sqrt {{x^2} + 1}$     AG

METHOD 2

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

$\sqrt {{x^2} + 1} \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y} \right)$     M1A1

Note:     Award M1 for attempting to express in the form $\sqrt {{x^2} + 1}  \times {\text{(LHS of de)}}$.

so $\sqrt {{x^2} + 1}$ is an integrating factor for this differential equation     AG

[4 marks]

$\sqrt {{x^2} + 1} \frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{\sqrt {{x^2} + 1} }}y = x\sqrt {{x^2} + 1}$ (or equivalent)     (M1)

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

$y\sqrt {{x^2} + 1}  = \int {x\sqrt {{x^2} + 1} {\text{d}}x{\text{ }}\left( {y = \frac{1}{{\sqrt {{x^2} + 1} }}\int {x\sqrt {{x^2} + 1} {\text{d}}x} } \right)}$     A1

$= \frac{1}{3}{({x^2} + 1)^{\frac{3}{2}}} + C$     (M1)A1

Note:     Award M1 for using an appropriate substitution.

Note:     Condone the absence of $C$.

substituting $x = 0,{\text{ }}y = 1 \Rightarrow C = \frac{2}{3}$     M1

Note:     Award M1 for attempting to find their value of $C$.

$y = \frac{1}{3}({x^2} + 1) + \frac{2}{{3\sqrt {{x^2} + 1} }}{\text{ }}\left( {y = \frac{{{{({x^2} + 1)}^{\frac{3}{2}}} + 2}}{{3\sqrt {{x^2} + 1} }}} \right)$     A1

[6 marks]