17N.3.AHL.TZ0.HCA_2

pestleMathematicsAAHLPaper 317N· ahl-5-18-1st-order-des-euler-method-variables-separable-integrating-factor-homogeneous-de-using-sub-y=vxsource ↗

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y = x$ where $y = 1$ when $x = 0$ .

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

[4]

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

[6]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]