Standard +0.3 This is a straightforward separable variables question requiring standard separation, integration using partial fractions (or recognizing the standard integral of 1/(1-y²)), and applying initial conditions. While it involves multiple steps, each is routine for A-level—slightly easier than average due to the clean setup and standard techniques involved.
6 The variables \(x\) and \(y\) are related by the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 - y ^ { 2 }$$
When \(x = 2 , y = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
Separate variables correctly and attempt integration of one side
B1
Obtain term \(\ln x\)
B1
State or imply \(\frac{1}{1-y^2} = \frac{A}{1-y} + \frac{B}{1+y}\) and use a relevant method to find \(A\) or \(B\)
M1
Obtain \(A = \frac{1}{2}, B = \frac{1}{2}\)
Integrate and obtain \(-\frac{1}{2}\ln(1-y) + \frac{1}{2}\ln(1+y)\), or equivalent
A1 \(\checkmark\)
[If the integral is directly stated as \(k_1\ln\left(\frac{1+y}{1-y}\right)\) or \(k_2\ln\left(\frac{1-y}{1+y}\right)\) give M1, and then A2 for \(k_1 = \frac{1}{2}\) or \(k_2 = -\frac{1}{2}\)]
Answer
Marks
Guidance
Evaluate a constant, or use limits \(x = 2, y = 0\) in a solution containing terms \(a\ln x, b\ln(1-y)\) and \(c\ln(1+y)\), where \(abc \neq 0\)
M1
Obtain solution in any correct form, e.g. \(\frac{1}{2}\ln\left(\frac{1+y}{1-y}\right) = \ln x - \ln 2\)
A1
Rearrange and obtain \(y = \frac{x^2-4}{x^2+4}\) or equivalent, free of logarithms
A1
[8]
Separate variables correctly and attempt integration of one side | B1 |
Obtain term $\ln x$ | B1 |
State or imply $\frac{1}{1-y^2} = \frac{A}{1-y} + \frac{B}{1+y}$ and use a relevant method to find $A$ or $B$ | M1 |
Obtain $A = \frac{1}{2}, B = \frac{1}{2}$ | |
Integrate and obtain $-\frac{1}{2}\ln(1-y) + \frac{1}{2}\ln(1+y)$, or equivalent | A1 $\checkmark$ |
[If the integral is directly stated as $k_1\ln\left(\frac{1+y}{1-y}\right)$ or $k_2\ln\left(\frac{1-y}{1+y}\right)$ give M1, and then A2 for $k_1 = \frac{1}{2}$ or $k_2 = -\frac{1}{2}$]
Evaluate a constant, or use limits $x = 2, y = 0$ in a solution containing terms $a\ln x, b\ln(1-y)$ and $c\ln(1+y)$, where $abc \neq 0$ | M1 |
Obtain solution in any correct form, e.g. $\frac{1}{2}\ln\left(\frac{1+y}{1-y}\right) = \ln x - \ln 2$ | A1 |
Rearrange and obtain $y = \frac{x^2-4}{x^2+4}$ or equivalent, free of logarithms | A1 | [8]
6 The variables $x$ and $y$ are related by the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 - y ^ { 2 }$$
When $x = 2 , y = 0$. Solve the differential equation, obtaining an expression for $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE P3 2012 Q6 [8]}}