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Date November 2017 Marks available 4 Reference code 17N.3ca.hl.TZ0.2
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Show that Question number 2 Adapted from N/A

Question

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]
a.

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

[6]
b.

Markscheme

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]

a.

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

b.

Examiners report

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a.
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b.

Syllabus sections

Topic 9 - Option: Calculus » 9.5
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