Edexcel FP2 2003 June — Question 12 10 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2003
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeHomogeneous equation (y = vx substitution)
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring homogeneous DE substitution, separable variables, and algebraic manipulation to reach a specific form. While the techniques are standard for FP2, the question demands careful execution across three connected parts with non-trivial algebra, placing it moderately above average difficulty.
Spec4.10a General/particular solutions: of differential equations

12. (a) Use the substitution \(y = v x\) to transform the equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + y ) ( x + y ) } { x ^ { 2 } } , x > 0$$ into the equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( 2 + v ) ^ { 2 }$$ (b) Solve the differential equation II to find \(\boldsymbol { v }\) as a function of \(\boldsymbol { x }\) (c) Hence show that \(\quad y = - 2 x - \frac { x } { \ln x + c }\), where \(c\) is an arbitrary constant, is a general solution of the differential equation I.

AnswerMarks Guidance
(a) \(y + x\frac{dv}{dx} = (4 + v)(1 + v)\)M1, M1
\(x\frac{dv}{dx} = v^2 + 5v + 4 - v\)A1
\(x\frac{dv}{dx} = (v + 2)^2\) \(\ast\)A1 (4)
(b) \(\int \frac{1}{(v+2)^2}dv = \int \frac{1}{x}dx\)B1, M1
\(-\frac{1}{2+v} = \ln x + c\)M1 A1 must have + c
\(2 + v = -\frac{1}{\ln x + c}\)M1
\(y = -\frac{1}{\ln x + c} - 2\)A1 (5)
(c) \(y = -2x - \frac{x}{\ln x + c}\)B1 (1)
(10 marks)
**(a)** $y + x\frac{dv}{dx} = (4 + v)(1 + v)$ | M1, M1 |

$x\frac{dv}{dx} = v^2 + 5v + 4 - v$ | A1 |

$x\frac{dv}{dx} = (v + 2)^2$ $\ast$ | A1 | (4) |

**(b)** $\int \frac{1}{(v+2)^2}dv = \int \frac{1}{x}dx$ | B1, M1 |

$-\frac{1}{2+v} = \ln x + c$ | M1 A1 | must have + c |

$2 + v = -\frac{1}{\ln x + c}$ | M1 |

$y = -\frac{1}{\ln x + c} - 2$ | A1 | (5) |

**(c)** $y = -2x - \frac{x}{\ln x + c}$ | B1 | (1) |

| | (10 marks) |

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12. (a) Use the substitution $y = v x$ to transform the equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + y ) ( x + y ) } { x ^ { 2 } } , x > 0$$

into the equation

$$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( 2 + v ) ^ { 2 }$$

(b) Solve the differential equation II to find $\boldsymbol { v }$ as a function of $\boldsymbol { x }$\\
(c) Hence show that $\quad y = - 2 x - \frac { x } { \ln x + c }$, where $c$ is an arbitrary constant, is a general solution of the differential equation I.\\

\hfill \mbox{\textit{Edexcel FP2 2003 Q12 [10]}}