| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Moderate -0.3 This is a straightforward separable variables question requiring standard manipulation (separating to get e^(-y)dy = xe^(-x)dx), integration by parts for the right side, and substituting initial conditions. Part (b) is routine substitution and logarithm manipulation. Slightly easier than average due to being a textbook-standard separable DE with clear steps. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct separation of variables: \(\int e^{-y}dy = \int xe^{-x}dx\) | B1 | Condone missing integral signs |
| Obtain term \(-e^{-y}\) | B1 | |
| Commence integration by parts and reach \(\pm xe^{-x} \pm \int e^{-x}dx\) | \*M1 | M0 if clearly using differentiation of a product |
| Complete integration and obtain \(-xe^{-x} - e^{-x}\) | A1 | |
| Use \(x=0\) and \(y=0\) to evaluate constant, in solution containing terms \(ae^{-y}, bxe^{-x}\) and \(ce^{-x}\), where \(abc \neq 0\) | DM1 | Must see working; should have \(-e^{-y}+C=-xe^{-x}-e^{-x}\) or equivalent |
| Correct solution in any form, must follow from correct working | A1 | e.g. \(-e^{-y}=-xe^{-x}-e^{-x}\); A0 if constant ignored or assumed zero |
| Obtain final answer \(y=-\ln\!\big((x+1)e^{-x}\big)\) from correct working | A1 | OE e.g. \(y=x-\ln(x+1)\), \(y=\ln\!\left(\frac{e^x}{x+1}\right)\); A0 if constant ignored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain answer \(y=1-\ln 2\) | B1 | Must follow from at least 6 or 7 marks obtained in part 6(a) |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct separation of variables: $\int e^{-y}dy = \int xe^{-x}dx$ | B1 | Condone missing integral signs |
| Obtain term $-e^{-y}$ | B1 | |
| Commence integration by parts and reach $\pm xe^{-x} \pm \int e^{-x}dx$ | \*M1 | M0 if clearly using differentiation of a product |
| Complete integration and obtain $-xe^{-x} - e^{-x}$ | A1 | |
| Use $x=0$ and $y=0$ to evaluate constant, in solution containing terms $ae^{-y}, bxe^{-x}$ and $ce^{-x}$, where $abc \neq 0$ | DM1 | Must see working; should have $-e^{-y}+C=-xe^{-x}-e^{-x}$ or equivalent |
| Correct solution in any form, must follow from correct working | A1 | e.g. $-e^{-y}=-xe^{-x}-e^{-x}$; A0 if constant ignored or assumed zero |
| Obtain final answer $y=-\ln\!\big((x+1)e^{-x}\big)$ from correct working | A1 | OE e.g. $y=x-\ln(x+1)$, $y=\ln\!\left(\frac{e^x}{x+1}\right)$; A0 if constant ignored |
**Total: 7 marks**
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain answer $y=1-\ln 2$ | B1 | Must follow from at least 6 or 7 marks obtained in part 6(a) |
**Total: 1 mark**
---6 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { y - x } ,$$
and $y = 0$ when $x = 0$.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation, obtaining an expression for $y$ in terms of $x$.
\item Find the value of $y$ when $x = 1$, giving your answer in the form $a - \ln b$, where $a$ and $b$ are integers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q6 [8]}}