Moderate -0.3 This is a straightforward separable variables question requiring standard technique: separate variables to get dy/y = (2-x²)/x dx, integrate both sides (requiring partial fractions or recognizing 2/x - x), then apply initial condition. While it requires competent algebraic manipulation, it's a routine textbook exercise with no novel insight needed, making it slightly easier than average.
5 The coordinates \(( x , y )\) of a general point on a curve satisfy the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 2 - x ^ { 2 } \right) y$$
The curve passes through the point \(( 1,1 )\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\).
Separate variables correctly and integrate at least one side
B1
Obtain term \(\ln y\)
B1
Obtain terms \(2\ln x - \frac{1}{2}x^2\)
B1+B1
Use \(x = 1\), \(y = 1\) to evaluate a constant, or as limits
M1
Obtain correct solution in any form, e.g. \(\ln y = 2\ln x - \frac{1}{2}x^2 + \frac{1}{2}\)
A1
Rearrange as \(y = x^2\exp\!\left(\frac{1}{2} - \frac{1}{2}x^2\right)\), or equivalent
A1
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and integrate at least one side | B1 | |
| Obtain term $\ln y$ | B1 | |
| Obtain terms $2\ln x - \frac{1}{2}x^2$ | B1+B1 | |
| Use $x = 1$, $y = 1$ to evaluate a constant, or as limits | M1 | |
| Obtain correct solution in any form, e.g. $\ln y = 2\ln x - \frac{1}{2}x^2 + \frac{1}{2}$ | A1 | |
| Rearrange as $y = x^2\exp\!\left(\frac{1}{2} - \frac{1}{2}x^2\right)$, or equivalent | A1 | |
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5 The coordinates $( x , y )$ of a general point on a curve satisfy the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 2 - x ^ { 2 } \right) y$$
The curve passes through the point $( 1,1 )$. Find the equation of the curve, obtaining an expression for $y$ in terms of $x$.\\
\hfill \mbox{\textit{CAIE P3 2018 Q5 [7]}}