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.
Solve the differential equation giving your answer in the form \(y = f(x)\).
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]
\(\sqrt {{x^2} + 1} \frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{\sqrt {{x^2} + 1} }}y = x\sqrt {{x^2} + 1} \) (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]