CAIE P3 2016 March — Question 7 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2016
SessionMarch
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (polynomial/exponential x-side)
DifficultyStandard +0.3 This is a straightforward separable variables question requiring standard techniques: separate to get e^(-y)dy = xe^x dx, integrate both sides (RHS needs integration by parts), apply initial condition, and rearrange for y. Part (ii) requires recognizing the domain restriction from ln(1-e^x), which is a simple observation once the solution is obtained. Slightly above average due to the integration by parts and final rearrangement, but still a standard textbook exercise.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

7 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { x + y }$$ and it is given that \(y = 0\) when \(x = 0\).
  1. Solve the differential equation and obtain an expression for \(y\) in terms of \(x\).
  2. Explain briefly why \(x\) can only take values less than 1 .

AnswerMarks Guidance
(i) Separate variables and attempt integration of one sideM1
Obtain term \(-e^{-y}\)A1
Integrate \(xe^x\) by parts reaching \(xe^{\pm}\int e^x \, dx\)M1
Obtain integral \(xe^x - e^x\)A1
Evaluate a constant, or use limits \(x = 0, y = 0\)M1
Obtain correct solution in any formA1
Obtain final answer \(y = -\ln(e^x(1-x))\), or equivalentA1 [7]
(ii) Justify the given statementB1 [1]
(i) Separate variables and attempt integration of one side | M1 |
Obtain term $-e^{-y}$ | A1 |
Integrate $xe^x$ by parts reaching $xe^{\pm}\int e^x \, dx$ | M1 |
Obtain integral $xe^x - e^x$ | A1 |
Evaluate a constant, or use limits $x = 0, y = 0$ | M1 |
Obtain correct solution in any form | A1 |
Obtain final answer $y = -\ln(e^x(1-x))$, or equivalent | A1 | [7]

(ii) Justify the given statement | B1 | [1]
7 The variables $x$ and $y$ satisfy the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { x + y }$$

and it is given that $y = 0$ when $x = 0$.\\
(i) Solve the differential equation and obtain an expression for $y$ in terms of $x$.\\
(ii) Explain briefly why $x$ can only take values less than 1 .

\hfill \mbox{\textit{CAIE P3 2016 Q7 [8]}}