Moderate -0.5 This is a straightforward separable variables question requiring separation, integration of standard functions (polynomial and exponential), and application of two boundary conditions to find constants. The integration is routine and the algebra is uncomplicated, making it slightly easier than average for A-level.
6 The variables \(x\) and \(y\) satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = k y ^ { 3 } \mathrm { e } ^ { - x }$$
where \(k\) is a constant. It is given that \(y = 1\) when \(x = 0\), and that \(y = \sqrt { } \mathrm { e }\) when \(x = 1\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
Separate variables correctly and attempt integration of at least one side
B1
Obtain term \(-\frac{1}{2y^2}\), or equivalent
B1
Obtain term \(-ke^{-x}\)
B1
Use a pair of limits, e.g. \(x=0\), \(y=1\) to obtain an equation in \(k\) and an arbitrary constant \(c\)
M1
Use a second pair of limits, e.g. \(x=1\), \(y=\sqrt{e}\), to obtain a second equation and solve for \(k\) or \(c\)
M1
Obtain \(k=\frac{1}{2}\) and \(c=0\)
A1
Obtain final answer \(y=e^{\frac{1}{2}x}\), or equivalent
A1
## Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and attempt integration of at least one side | B1 | |
| Obtain term $-\frac{1}{2y^2}$, or equivalent | B1 | |
| Obtain term $-ke^{-x}$ | B1 | |
| Use a pair of limits, e.g. $x=0$, $y=1$ to obtain an equation in $k$ and an arbitrary constant $c$ | M1 | |
| Use a second pair of limits, e.g. $x=1$, $y=\sqrt{e}$, to obtain a second equation and solve for $k$ or $c$ | M1 | |
| Obtain $k=\frac{1}{2}$ and $c=0$ | A1 | |
| Obtain final answer $y=e^{\frac{1}{2}x}$, or equivalent | A1 | |
6 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = k y ^ { 3 } \mathrm { e } ^ { - x }$$
where $k$ is a constant. It is given that $y = 1$ when $x = 0$, and that $y = \sqrt { } \mathrm { e }$ when $x = 1$. Solve the differential equation, obtaining an expression for $y$ in terms of $x$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q6 [7]}}