Pre-U Pre-U 9795/1 2016 Specimen — Question 9 13 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2016
SessionSpecimen
Marks13
TopicFirst order differential equations (integrating factor)
TypeBernoulli equation
DifficultyChallenging +1.2 This is a guided Bernoulli equation problem where part (i) walks students through the substitution, removing the main conceptual hurdle. Part (ii) requires applying integrating factor method and using initial conditions—standard Further Maths techniques with moderate algebraic manipulation but no novel insight required.
Spec4.10a General/particular solutions: of differential equations4.10c 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)
\(\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4\) is a *Bernoulli (differential) equation*
\(u = \frac{1}{y^3} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{3}{y^4} \times \frac{\mathrm{d}y}{\mathrm{d}x}\) B1
Then \(\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4\) becomes \(-\frac{3}{y^4}\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{3}{y^3} = -9x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} - 3u = -9x\) AG M1 A1
(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
General solution 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
Method 2
Auxiliary equation \(m - 3 = 0 \Rightarrow u_C = A\mathrm{e}^{3x}\) is the complementary function M1 A1
For particular integral try \(u_P = ax + b\), \(u_P' = a\) M1
Substituting \(u_P = ax+b\) and \(u_P' = a\) into the d.e. and comparing terms M1
\(a - 3ax - 3b = -9x \Rightarrow a = 3,\, b = 1\) i.e. \(u_P = 3x+1\) A1
General solution is \(u = 3x + 1 + A\mathrm{e}^{3x}\) ft particular integral + complementary function B1
\(\Rightarrow y^{-3} = \frac{1}{3x+1+A\mathrm{e}^{3x}}\) ft B1
Using \(x=0\), \(y = \frac{1}{2}\) to find \(A\): \(A = 7\) or \(y^3 = \frac{1}{3x+1+7\mathrm{e}^{3x}}\) M1 A1
Total: 13 marks
**(i)**

$\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4$ is a *Bernoulli (differential) equation*

$u = \frac{1}{y^3} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{3}{y^4} \times \frac{\mathrm{d}y}{\mathrm{d}x}$ **B1**

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

**(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**

General solution 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**

**Method 2**

Auxiliary equation $m - 3 = 0 \Rightarrow u_C = A\mathrm{e}^{3x}$ is the complementary function **M1 A1**

For particular integral try $u_P = ax + b$, $u_P' = a$ **M1**

Substituting $u_P = ax+b$ and $u_P' = a$ into the d.e. and comparing terms **M1**

$a - 3ax - 3b = -9x \Rightarrow a = 3,\, b = 1$ i.e. $u_P = 3x+1$ **A1**

General solution is $u = 3x + 1 + A\mathrm{e}^{3x}$ **ft** particular integral + complementary function **B1**

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

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

**Total: 13 marks**
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 2016 Q9 [13]}}
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