| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.3 This is a straightforward separable variables question requiring standard technique (separating variables, integrating both sides including integration by parts for xe^x, applying initial conditions). Part (ii) adds mild analysis of the domain from the solution form. Slightly easier than average due to being a textbook application with clear method, though the integration by parts and domain analysis provide some substance. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Separate variables correctly and attempt integration of at least one side | B1 | \(\int e^{-y}\,dy = \int xe^x\,dx\) |
| Obtain term \(-e^{-y}\) | B1 | B0B1 is possible |
| Commence integration by parts and reach \(xe^x \pm \int e^x\,dx\) | M1 | B0B0M1A1 is possible |
| Obtain \(xe^x - e^x\) | A1 | or equivalent; B1B1M1A1 available if no constant of integration |
| Use \(x=0,\, y=0\) to evaluate constant, or as limits in definite integral, in solution with terms \(ae^{-y}\), \(bxe^x\) and \(ce^x\), where \(abc \neq 0\) | M1 | Must see this step |
| Obtain correct solution in any form | A1 | e.g. \(e^{-y} = e^x - xe^x\) |
| Rearrange as \(y = -\ln(1-x) - x\) | A1 | or equivalent e.g. \(y = \ln\dfrac{1}{e^x(1-x)}\); ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Justify the given statement | B1 | e.g. require \(1-x > 0\) for the \(\ln\) term to exist, hence \(x < 1\). Must be considering range of values of \(x\), and must be relevant to *their* \(y\) involving \(\ln(1-x)\) |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Separate variables correctly and attempt integration of at least one side | B1 | $\int e^{-y}\,dy = \int xe^x\,dx$ |
| Obtain term $-e^{-y}$ | B1 | B0B1 is possible |
| Commence integration by parts and reach $xe^x \pm \int e^x\,dx$ | M1 | B0B0M1A1 is possible |
| Obtain $xe^x - e^x$ | A1 | or equivalent; B1B1M1A1 available if no constant of integration |
| Use $x=0,\, y=0$ to evaluate constant, or as limits in definite integral, in solution with terms $ae^{-y}$, $bxe^x$ and $ce^x$, where $abc \neq 0$ | M1 | Must see this step |
| Obtain correct solution in any form | A1 | e.g. $e^{-y} = e^x - xe^x$ |
| Rearrange as $y = -\ln(1-x) - x$ | A1 | or equivalent e.g. $y = \ln\dfrac{1}{e^x(1-x)}$; ISW |
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## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Justify the given statement | B1 | e.g. require $1-x > 0$ for the $\ln$ term to exist, hence $x < 1$. Must be considering range of values of $x$, and must be relevant to *their* $y$ involving $\ln(1-x)$ |
---7 The variables $x$ and $y$ satisfy the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { x + y }$. It is given that $y = 0$ when $x = 0$.\\
(i) Solve the differential equation, obtaining $y$ in terms of $x$.\\
(ii) Explain why $x$ can only take values that are less than 1 .\\
\hfill \mbox{\textit{CAIE P3 2019 Q7 [8]}}