Challenging +1.2 This is a standard homogeneous differential equation requiring the y=vx substitution (given), separation of variables, and partial fractions. While it involves multiple techniques and algebraic manipulation across three parts, the method is completely prescribed and follows a textbook template for FP2. The algebra is moderately involved but routine for Further Maths students.
3. (a) Show that the substitution \(y = v x\) transforms the differential equation
$$\frac { d y } { d x } = \frac { 3 x - 4 y } { 4 x + 3 y }$$
into the differential equation
$$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \frac { 3 v ^ { 2 } + 8 v - 3 } { 3 v + 4 }$$
(b) By solving differential equation (II), find a general solution of differential equation (I). (5)
(c) Given that \(y = 7\) at \(x = 1\), show that the particular solution of differential equation (I) can be written as
$$( 3 y - x ) ( y + 3 x ) = 200$$
(5)(Total 14 marks)
Removing ln's correctly at any stage, dep. on having \(C\)
Using \((1, 7)\) to form an equation in \(A\)
M1
Need not be \(A = ...\)
\((1,7) \Rightarrow 3 \times 49 + 56 - 3 = A \Rightarrow A = 200\)
(or equiv., can still be ln)A1
\(3y^2 + 8yx - 3x^2 = 200\)
\((3y - x)(y + 3x) = 200\)
M1, A1
M dependent on 2 previous M's) M1 A1 eso 5
Note: Parts (b) and (c) may well merge. Alternative (b): Partial fractions may be used \(\left(A = \frac{3}{2}, B = \frac{1}{2}\right)\), giving \(\frac{1}{2}\ln(3y - 1) + \frac{1}{2}\ln(v + 3)\). Final M in (c) requires formation and factorisation of the quadratic.
**Part (a)**
| $\frac{dy}{dx} = v + x\frac{dv}{dx}$ | B1 | |
| $v + x\frac{dv}{dx} = \frac{3x - 4ux}{4x + 3vx}$ (all in terms of $v$ and $x$) | M1 | |
| $x\frac{dv}{dx} = \frac{3 - 4v - v(4 + 3v)}{4 + 3v}$ | M1 | Requires $\frac{dv}{dx} = f(v)$, 2 terms over common denom. |
| $x\frac{dv}{dx} = \frac{3v^2 + 8v - 3}{3v + 4}$ | A1 | eso 4 |
**Part (b)**
| $\frac{3v + 4}{3v^2 + 8v - 3}dv = -\frac{1}{x}dx$ | M1 | Separating variables |
| $\pm \ln x$ | B1 | |
| $\frac{1}{2}\ln(3v^2 + 8v - 3)$ | M1, A1 | M: $k\ln(3v^2 + 8v - 3)$ |
| $\frac{1}{2}\ln\left(\frac{3y^2}{x^2} + \frac{8y}{x} - 3\right) = -\ln x + C$ | A1 | Or any equivalent form 5 |
**Part (c)**
| $\frac{3y^2}{x^2} + \frac{8y}{x} - 3 = \frac{A}{x^2}$ | M1 | Removing ln's correctly at any stage, dep. on having $C$ |
| Using $(1, 7)$ to form an equation in $A$ | M1 | Need not be $A = ...$ |
| $(1,7) \Rightarrow 3 \times 49 + 56 - 3 = A \Rightarrow A = 200$ | | (or equiv., can still be ln)A1 |
| $3y^2 + 8yx - 3x^2 = 200$ | | |
| $(3y - x)(y + 3x) = 200$ | M1, A1 | M dependent on 2 previous M's) M1 A1 eso 5 |
Note: Parts (b) and (c) may well merge. Alternative (b): Partial fractions may be used $\left(A = \frac{3}{2}, B = \frac{1}{2}\right)$, giving $\frac{1}{2}\ln(3y - 1) + \frac{1}{2}\ln(v + 3)$. Final M in (c) requires formation and factorisation of the quadratic.
3. (a) Show that the substitution $y = v x$ transforms the differential equation
$$\frac { d y } { d x } = \frac { 3 x - 4 y } { 4 x + 3 y }$$
into the differential equation
$$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \frac { 3 v ^ { 2 } + 8 v - 3 } { 3 v + 4 }$$
(b) By solving differential equation (II), find a general solution of differential equation (I). (5)\\
(c) Given that $y = 7$ at $x = 1$, show that the particular solution of differential equation (I) can be written as
$$( 3 y - x ) ( y + 3 x ) = 200$$
(5)(Total 14 marks)\\
\hfill \mbox{\textit{Edexcel FP2 2006 Q3 [14]}}