Pre-U Pre-U 9795/1 2012 June — Question 9 9 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2012
SessionJune
Marks9
TopicFirst order differential equations (integrating factor)
TypeBernoulli equation
DifficultyChallenging +1.2 This is a structured Bernoulli equation problem where part (i) guides students through the substitution, removing the main conceptual hurdle. Part (ii) requires applying an integrating factor and back-substitution—standard Further Maths techniques. The scaffolding in part (i) significantly reduces difficulty compared to an unguided Bernoulli equation, placing this moderately above average difficulty.
Spec4.10c Integrating factor: first order equations

9
  1. Show that the substitution \(u = \frac { 1 } { y ^ { 3 } }\) transforms the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\) into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x$$
  2. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\), given that \(y = \frac { 1 } { 2 }\) when \(x = 0\). Give your answer in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

(i) \(u = \frac{1}{y^3} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{3}{y^4}\times\frac{\mathrm{d}y}{\mathrm{d}x}\) B1
Then \(\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4\) becomes \(-\frac{3}{y^4}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{3}{y^3} = -9x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} - 3u = -9x\) AG M1 A1
[3]
(ii) Method 1
IF is \(\mathrm{e}^{-3x}\) M1 A1
\(\Rightarrow u\mathrm{e}^{-3x} = \int -9x\mathrm{e}^{-3x}\,\mathrm{d}x\) M1
\(= 3x\mathrm{e}^{-3x} - \int 3\mathrm{e}^{-3x}\,\mathrm{d}x\) Use of "parts" M1
\(= (3x+1)\mathrm{e}^{-3x} + C\) A1
Gen. Soln. is \(u = 3x + 1 + C\mathrm{e}^{3x}\) ft B1
\(\Rightarrow y^3 = \frac{1}{3x+1+C\mathrm{e}^{3x}}\) ft B1
Using \(x = 0\), \(y = \frac{1}{2}\) to find \(C\): \(C = 7\) or \(y^3 = \frac{1}{3x+1+7\mathrm{e}^{3x}}\) M1 A1
[9]
**(i)** $u = \frac{1}{y^3} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{3}{y^4}\times\frac{\mathrm{d}y}{\mathrm{d}x}$ **B1**

Then $\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4$ becomes $-\frac{3}{y^4}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{3}{y^3} = -9x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} - 3u = -9x$ **AG** **M1 A1**

**[3]**

**(ii) Method 1**

IF is $\mathrm{e}^{-3x}$ **M1 A1**

$\Rightarrow u\mathrm{e}^{-3x} = \int -9x\mathrm{e}^{-3x}\,\mathrm{d}x$ **M1**

$= 3x\mathrm{e}^{-3x} - \int 3\mathrm{e}^{-3x}\,\mathrm{d}x$ Use of "parts" **M1**

$= (3x+1)\mathrm{e}^{-3x} + C$ **A1**

Gen. Soln. is $u = 3x + 1 + C\mathrm{e}^{3x}$ **ft** **B1**

$\Rightarrow y^3 = \frac{1}{3x+1+C\mathrm{e}^{3x}}$ **ft** **B1**

Using $x = 0$, $y = \frac{1}{2}$ to find $C$: $C = 7$ or $y^3 = \frac{1}{3x+1+7\mathrm{e}^{3x}}$ **M1 A1**

**[9]**
9 (i) Show that the substitution $u = \frac { 1 } { y ^ { 3 } }$ transforms the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }$ into

$$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x$$

(ii) Solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }$, given that $y = \frac { 1 } { 2 }$ when $x = 0$. Give your answer in the form $y ^ { 3 } = \mathrm { f } ( x )$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2012 Q9 [9]}}