Standard +0.3 This is a standard linear first-order differential equation requiring the integrating factor method. While it's from Further Maths F2, the technique is routine: divide by x to get standard form, find integrating factor e^(∫(1-2x)/x dx) = e^(ln x - 2x) = xe^(-2x), then integrate. The algebra is straightforward with no conceptual surprises, making it slightly easier than an average A-level question despite being Further Maths content.
6.
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( 1 - 2 x ) y = x , \quad x > 0$$
Find the general solution of the differential equation, giving your answer in the form \(y = \mathrm { f } ( x )\).
Must have \(y=\ldots\) Must include a constant. ft their previous line
Total: 9
## Question 6:
$x\frac{dy}{dx} + (1-2x)y = x$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} + \frac{(1-2x)}{x}y (=1)$ | M1 | Divides by $x$ - may be implied by subsequent working |
| Integrating factor $I = e^{\int\frac{1-2x}{x}dx}$ | dM1 | Correct attempt at $I$, including an attempt at the integration. ln must be seen if not fully correct |
| $= e^{\ln(x)-2x}$ | A1 | Correct expression |
| $= xe^{-2x}$ | A1 | No errors in working allowed |
| $xye^{-2x} = \int xe^{-2x}dx$ | M1 | Multiply through by their IF and integrate LHS. Can be given if $y \times$ their IF is seen |
| $= -\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x}dx$ | M1A1 | M1: Correct application by parts ie differentiate $x$ and attempt to integrate $e^{-2x}$. A1: Correct expression |
| $= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\ (+c)$ | A1 | Complete the integration to a correct result. Constant not required |
| $y = \frac{ce^{2x}}{x} - \frac{1}{4x} - \frac{1}{2}$ | A1ft | Must have $y=\ldots$ Must include a constant. ft their previous line |
**Total: 9**
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6.
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( 1 - 2 x ) y = x , \quad x > 0$$
Find the general solution of the differential equation, giving your answer in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel F2 2014 Q6 [9]}}