| Date | None Specimen | Marks available | 7 | Reference code | SPNone.3ca.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
| Command term | Find | Question number | 2 | Adapted from | N/A |
Question
Consider the differential equation
\[x\frac{{{\text{d}}y}}{{{\text{d}}x}} = y + \sqrt {{x^2} - {y^2}} ,{\text{ }}x > 0,{\text{ }}{x^2} > {y^2}.\]
Show that this is a homogeneous differential equation.
Find the general solution, giving your answer in the form \(y = f(x)\) .
Markscheme
the equation can be rewritten as
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y + \sqrt {{x^2} - {y^2}} }}{x} = \frac{y}{x} + \sqrt {1 - {{\left( {\frac{y}{x}} \right)}^2}} \) A1
so the differential equation is homogeneous AG
[1 mark]
put y = vx so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\) M1A1
substituting,
\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v + \sqrt {1 - {v^2}} \) M1
\(\int {\frac{{{\text{d}}v}}{{\sqrt {1 - {v^2}} }} = \int {\frac{{{\text{d}}x}}{x}} } \) M1
\(\arcsin v = \ln x + C\) A1
\(\frac{y}{x} = \sin (\ln x + C)\) A1
\(y = x\sin (\ln x + C)\) A1
[7 marks]
