Moderate -0.8 This is a straightforward separable variables question requiring only basic manipulation (rewrite as e^(2x)·e^y, separate to dy/e^y = e^(2x)dx, integrate both sides to get -e^(-y) = (1/2)e^(2x) + c, then apply initial condition). It's more routine than the average A-level question since it follows a standard template with no conceptual obstacles, though it does require careful algebraic manipulation to reach the final form for y.
5 The variables \(x\) and \(y\) satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x + y }$$
and \(y = 0\) when \(x = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
Separate variables correctly and attempt integration of both sides
B1
Obtain term \(-e^{-y}\), or equivalent
B1
Obtain term \(\frac{1}{2}e^{2x}\), or equivalent
B1
Evaluate a constant, or use limits \(x = 0, y = 0\) in a solution containing terms \(ae^{-y}\) and \(be^{2x}\)
M1
Obtain correct solution in any form, e.g. \(-e^{-y} = \frac{1}{2}e^{2x} - \frac{3}{2}\)
A1
Rearrange and obtain \(y = \ln(2/(3-e^{2x}))\), or equivalent
A1
[6]
| Separate variables correctly and attempt integration of both sides | B1 | |
| Obtain term $-e^{-y}$, or equivalent | B1 | |
| Obtain term $\frac{1}{2}e^{2x}$, or equivalent | B1 | |
| Evaluate a constant, or use limits $x = 0, y = 0$ in a solution containing terms $ae^{-y}$ and $be^{2x}$ | M1 | |
| Obtain correct solution in any form, e.g. $-e^{-y} = \frac{1}{2}e^{2x} - \frac{3}{2}$ | A1 | |
| Rearrange and obtain $y = \ln(2/(3-e^{2x}))$, or equivalent | A1 | [6] |
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5 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x + y }$$
and $y = 0$ when $x = 0$. Solve the differential equation, obtaining an expression for $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE P3 2012 Q5 [6]}}