Pre-U Pre-U 9795/1 2019 Specimen — Question 9 3 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2019
SessionSpecimen
Marks3
TopicFirst order differential equations (integrating factor)
TypeBernoulli equation
DifficultyChallenging +1.2 This is a guided Bernoulli equation problem where part (a) walks students through the substitution, reducing it to mechanical differentiation and algebraic manipulation. Part (b) requires solving a standard linear first-order ODE with integrating factor and applying initial conditions. While it involves multiple techniques (substitution, integrating factor, boundary conditions), the structure is highly scaffolded and follows a standard template taught in Further Maths courses, making it moderately above average difficulty but not requiring novel insight.
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 \(y ^ { 3 } = \mathrm { f } ( x )\).
(a) \(\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4\) is a *Bernoulli (differential) equation* B1
\(u = \frac{1}{y^3} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{3}{y^4}\times\frac{\mathrm{d}y}{\mathrm{d}x}\)
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 M1A1
Total: 3
(b) METHOD 1
Integrating factor is \(e^{\int -3\,\mathrm{d}x} = e^{-3x}\) M1A1
\(\Rightarrow ue^{-3x} = \int -9xe^{-3x}\,\mathrm{d}x\) M1
\(= 3xe^{-3x} - \int 3e^{-3x}\,\mathrm{d}x\) Use of "parts" M1
\(= (3x+1)e^{-3x} + C\) A1
General solution is \(u = 3x + 1 + Ce^{3x}\) ft B1 (B1ft)
\(\Rightarrow y^3 = \frac{1}{3x+1+Ce^{3x}}\) ft B1 (B1ft)
Using \(x=0\), \(y=\frac{1}{2}\) to find \(C\) M1
\(C = 7\) or \(y^3 = \frac{1}{3x+1+7e^{3x}}\) A1
OR METHOD 2
Auxiliary equation \(m - 3 = 0 \Rightarrow u_c = Ae^{3x}\) is the complementary function M1A1
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 \(u = 3x + 1 + Ae^{3x}\) ft particular integral + complementary function provided particular integral has no arbitrary constants and complementary function has one B1 (B1ft)
\(\Rightarrow y^3 = \frac{1}{3x+1+Ae^{3x}}\) ft B1 (B1ft)
Using \(x=0\), \(y=\frac{1}{2}\) to find \(A\) M1
\(A = 7\) or \(y^3 = \frac{1}{3x+1+7e^{3x}}\) A1
Total: 9
**(a)** $\frac{\mathrm{d}y}{\mathrm{d}x} + y = 3xy^4$ is a *Bernoulli (differential) equation* **B1**

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

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** **M1A1**

**Total: 3**

**(b) METHOD 1**

Integrating factor is $e^{\int -3\,\mathrm{d}x} = e^{-3x}$ **M1A1**

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

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

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

General solution is $u = 3x + 1 + Ce^{3x}$ **ft** **B1** (B1ft)

$\Rightarrow y^3 = \frac{1}{3x+1+Ce^{3x}}$ **ft** **B1** (B1ft)

Using $x=0$, $y=\frac{1}{2}$ to find $C$ **M1**

$C = 7$ or $y^3 = \frac{1}{3x+1+7e^{3x}}$ **A1**

**OR METHOD 2**

Auxiliary equation $m - 3 = 0 \Rightarrow u_c = Ae^{3x}$ is the complementary function **M1A1**

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 $u = 3x + 1 + Ae^{3x}$ **ft** particular integral + complementary function provided particular integral has no arbitrary constants and complementary function has one **B1** (B1ft)

$\Rightarrow y^3 = \frac{1}{3x+1+Ae^{3x}}$ **ft** **B1** (B1ft)

Using $x=0$, $y=\frac{1}{2}$ to find $A$ **M1**

$A = 7$ or $y^3 = \frac{1}{3x+1+7e^{3x}}$ **A1**

**Total: 9**
9
\begin{enumerate}[label=(\alph*)]
\item 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

$$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x$$
\item 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 $y ^ { 3 } = \mathrm { f } ( x )$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2019 Q9 [3]}}
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