Moderate -0.5 This is a straightforward separable variables question requiring only separation, integration of standard forms (∫y⁻¹dy and ∫x/(1+x²)dx), and application of initial conditions. The integration is routine (ln|y| and ½ln(1+x²)), and the algebra to solve for y is simple. Slightly easier than average as it's a textbook example with no complications.
4 The variables \(x\) and \(y\) satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x y } { 1 + x ^ { 2 } }$$
and \(y = 2\) when \(x = 0\).
Solve the differential equation, obtaining a simplified expression for \(y\) in terms of \(x\).
\(\int\frac{x}{1+x^2}dx=\int\frac{1}{y}dy\) Accept without integral signs
Obtain term \(\ln y\)
B1
State term of the form \(k\ln(1+x^2)\)
M1
State correct term \(\frac{1}{2}\ln(1+x^2)\)
A1
OE
Evaluate a constant, or use limits \(x=0\), \(y=2\) in a solution containing terms \(a\ln y\) and \(b\ln(1+x^2)\) where \(ab\neq 0\)
M1
If they remove logs first the constant must be of the correct form
Obtain correct solution in any form
A1
e.g. \(\ln y+\ln\frac{1}{2}=\frac{1}{2}\ln(1+x^2)\)
Simplify and obtain \(y=2\sqrt{1+x^2}\)
A1
OE. The question asks for simplification, so need to deal with \(\exp(\ln(...))\)
Total: 7 marks
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly | B1 | $\int\frac{x}{1+x^2}dx=\int\frac{1}{y}dy$ Accept without integral signs |
| Obtain term $\ln y$ | B1 | |
| State term of the form $k\ln(1+x^2)$ | M1 | |
| State correct term $\frac{1}{2}\ln(1+x^2)$ | A1 | OE |
| Evaluate a constant, or use limits $x=0$, $y=2$ in a solution containing terms $a\ln y$ and $b\ln(1+x^2)$ where $ab\neq 0$ | M1 | If they remove logs first the constant must be of the correct form |
| Obtain correct solution in any form | A1 | e.g. $\ln y+\ln\frac{1}{2}=\frac{1}{2}\ln(1+x^2)$ |
| Simplify and obtain $y=2\sqrt{1+x^2}$ | A1 | OE. The question asks for simplification, so need to deal with $\exp(\ln(...))$ |
**Total: 7 marks**
---
4 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x y } { 1 + x ^ { 2 } }$$
and $y = 2$ when $x = 0$.\\
Solve the differential equation, obtaining a simplified expression for $y$ in terms of $x$.\\
\hfill \mbox{\textit{CAIE P3 2022 Q4 [7]}}