Standard +0.3 This is a standard A-level differential equations question that follows a predictable two-part structure: decompose into partial fractions, then integrate using separation of variables. Part (a) is routine partial fractions with linear factors. Part (b) requires recognizing the separable form and applying the result from (a), followed by finding a particular solution using initial conditions. While it involves multiple techniques, each step is algorithmic with no novel insight required, making it slightly easier than average.
6. (a) Express \(\frac { 5 - 4 x } { ( 2 x - 1 ) ( x + 1 ) }\) in partial fractions.
(b) (i) Find a general solution of the differential equation
$$( 2 x - 1 ) ( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 5 - 4 x ) y , \quad x > \frac { 1 } { 2 }$$
Given that \(y = 4\) when \(x = 2\),
(ii) find the particular solution of this differential equation. Give your answer in the form \(y = \mathrm { f } ( x )\).
RHS correct integration for their partial fractions – do not need LHS nor \(+c\)
\(\ln y = \ln(2x-1) - 3\ln(x+1) + c\)
A1
All three terms correct (LHS and RHS) including \(+c\)
\(\ln 4 = \ln(2(2)-1) - 3\ln(2+1) + c \Rightarrow c = \{\ln 36\}\)
M1
Substitutes \(y=4\) and \(x=2\) into their general solution with constant of integration to obtain \(c=\)
\(\ln y = \ln(2x-1) - 3\ln(x+1) + \ln 36\) so \(\ln y = \ln\left(\frac{36(2x-1)}{(x+1)^3}\right)\) so \(y = \frac{36(2x-1)}{(x+1)^3}\)
M1 A1
M1: Fully correct method of removing logs – must have constant of integration treated correctly. Must have had \(\ln y =\) .... earlier. A1: \(y = \frac{36(2x-1)}{(x+1)^3}\) isw
[7]
Method 2 for (ii):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Solution as Method 1 up to \(\ln y = \ln(2x-1) - 3\ln(x+1) + c\), so first four marks as before
B1M1A1A1
Writes \(y = \frac{A(2x-1)}{(x+1)^3}\) as general solution
M1
Earns 3rd M1 mark
Then substitutes to find constant \(A\)
M1
Earns 2nd M1 mark
\(y = \frac{36(2x-1)}{(x+1)^3}\)
A1
[7] Total: 10
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5-4x}{(2x-1)(x+1)} \equiv \frac{A}{(2x-1)} + \frac{B}{(x+1)}$, so $5-4x \equiv A(x+1) + B(2x-1)$ | B1 | Forming the linear identity (may be implied). Note: A & B are not assigned – other letters may be used |
| Let $x=-1$, $9=B(-3) \Rightarrow B=$ ... Let $x=\frac{1}{2}$, $3=A\left(\frac{3}{2}\right) \Rightarrow A=$ ... | M1 | Valid method to find value of one of either A or B |
| $A=2$ and $B=-3$ or $\left\{\frac{5-4x}{(2x-1)(x+1)} \equiv \frac{2}{(2x-1)} - \frac{3}{(x+1)}\right\}$ | A1 | Both values correct. Sufficient without rewriting answer provided it is clear what A and B are. Note: cover-up rule from no working is B1M1A1 |
| **[3]** | | |
### Part (b)(i) and (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{y}\, dy = \int \frac{5-4x}{(2x-1)(x+1)}\, dx$ | B1 | Separates variables as shown. Can be implied. Need both sides correct, but condone missing integral signs |
| $= \int \frac{2}{(2x-1)} - \frac{3}{(x+1)}\, dx = C\ln(2x-1) + D\ln(x+1)$ | M1 | Uses partial fractions on RHS and obtains two log terms after integration. Coefficients may be wrong. Ignore LHS for this mark |
| $= \frac{``2"}{2}\ln(2x-1) - \text{``}3\text{''}\ln(x+1)$ | A1ft | RHS correct integration for their partial fractions – do not need LHS nor $+c$ |
| $\ln y = \ln(2x-1) - 3\ln(x+1) + c$ | A1 | All three terms correct (LHS and RHS) including $+c$ |
| $\ln 4 = \ln(2(2)-1) - 3\ln(2+1) + c \Rightarrow c = \{\ln 36\}$ | M1 | Substitutes $y=4$ and $x=2$ into their general solution with constant of integration to obtain $c=$ |
| $\ln y = \ln(2x-1) - 3\ln(x+1) + \ln 36$ so $\ln y = \ln\left(\frac{36(2x-1)}{(x+1)^3}\right)$ so $y = \frac{36(2x-1)}{(x+1)^3}$ | M1 A1 | M1: Fully correct method of removing logs – must have constant of integration treated correctly. Must have had $\ln y =$ .... earlier. A1: $y = \frac{36(2x-1)}{(x+1)^3}$ isw |
| **[7]** | | |
**Method 2 for (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solution as Method 1 up to $\ln y = \ln(2x-1) - 3\ln(x+1) + c$, so first four marks as before | B1M1A1A1 | |
| Writes $y = \frac{A(2x-1)}{(x+1)^3}$ as general solution | M1 | Earns 3rd M1 mark |
| Then substitutes to find constant $A$ | M1 | Earns 2nd M1 mark |
| $y = \frac{36(2x-1)}{(x+1)^3}$ | A1 | |
| **[7] Total: 10** | | |
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6. (a) Express $\frac { 5 - 4 x } { ( 2 x - 1 ) ( x + 1 ) }$ in partial fractions.\\
(b) (i) Find a general solution of the differential equation
$$( 2 x - 1 ) ( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 5 - 4 x ) y , \quad x > \frac { 1 } { 2 }$$
Given that $y = 4$ when $x = 2$,\\
(ii) find the particular solution of this differential equation. Give your answer in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel C34 2014 Q6 [10]}}