| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation question with standard partial fractions decomposition. Part (i) is routine A-level algebra, and part (ii) follows a well-practiced method: separate variables, integrate both sides using the given partial fractions, and rearrange to the required form. The integration and algebraic manipulation are mechanical with no novel insight required, making this slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{(y-2)(y+1)} = \frac{A}{y-2} + \frac{B}{y+1} = \frac{A(y+1)+B(y-2)}{(y-2)(y+1)}\) | ||
| \(3 = A(y+1) + B(y-2)\) | M1 | Substituting, equating coeffs or cover up |
| \(y=2 \Rightarrow 3=3A \Rightarrow A=1\) | A1 | |
| \(y=-1 \Rightarrow 3=-3B \Rightarrow B=-1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = x^2(y-2)(y+1) \Rightarrow \int \frac{3\,dy}{(y-2)(y+1)} = \int 3x^2\,dx\) | M1 | Separating variables |
| \(\int\left(\frac{1}{y-2} - \frac{1}{y+1}\right)dy = \int 3x^2\,dx\) | ||
| \(\ln(y-2) - \ln(y+1) = x^3 + c\) | B1ft | \(\ln(y-2) - \ln(y+1)\), ft their \(A\), \(B\) |
| \(\ln\!\left(\frac{y-2}{y+1}\right) = x^3 + c\) | B1 | \(x^3 + c\) |
| \(\frac{y-2}{y+1} = e^{x^3+c} = e^{x^3}\cdot e^c = Ae^{x^3}\) | M1 | Anti-logging including \(c\) |
| E1 | www |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{(y-2)(y+1)} = \frac{A}{y-2} + \frac{B}{y+1} = \frac{A(y+1)+B(y-2)}{(y-2)(y+1)}$ | | |
| $3 = A(y+1) + B(y-2)$ | M1 | Substituting, equating coeffs or cover up |
| $y=2 \Rightarrow 3=3A \Rightarrow A=1$ | A1 | |
| $y=-1 \Rightarrow 3=-3B \Rightarrow B=-1$ | A1 | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = x^2(y-2)(y+1) \Rightarrow \int \frac{3\,dy}{(y-2)(y+1)} = \int 3x^2\,dx$ | M1 | Separating variables |
| $\int\left(\frac{1}{y-2} - \frac{1}{y+1}\right)dy = \int 3x^2\,dx$ | | |
| $\ln(y-2) - \ln(y+1) = x^3 + c$ | B1ft | $\ln(y-2) - \ln(y+1)$, ft their $A$, $B$ |
| $\ln\!\left(\frac{y-2}{y+1}\right) = x^3 + c$ | B1 | $x^3 + c$ |
| $\frac{y-2}{y+1} = e^{x^3+c} = e^{x^3}\cdot e^c = Ae^{x^3}$ | M1 | Anti-logging including $c$ |
| | E1 | www |7 (i) Express $\frac { 3 } { ( y - 2 ) ( y + 1 ) }$ in partial fractions.\\[0pt]
[3]\\
(ii) Hence, given that $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$
show that $\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }$, where $A$ is a constant.
\hfill \mbox{\textit{OCR MEI C4 Q7 [8]}}