Challenging +1.2 This is a Further Maths FP3 question requiring a guided substitution to convert a non-linear first-order ODE into a linear form solvable by integrating factor. While the substitution is provided (reducing problem-solving demand), students must correctly apply the chain rule, recognize the resulting linear form, and execute the integrating factor method—a multi-step process requiring solid technique and careful algebra that goes beyond routine A-level pure maths.
2 Use the substitution \(u = y ^ { 2 }\) to find the general solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = \frac { \mathrm { e } ^ { x } } { y }$$
for \(y\) in terms of \(x\).
Correctly finds. Or \(\dfrac{dy}{dx}=\dfrac{1}{2}u^{-\frac{1}{2}}\dfrac{du}{dx}\)
So DE \(\Rightarrow 2y\dfrac{dy}{dx}-4y^2=2e^x\)
M1
Or for complete unsimplified substitution
\(\Rightarrow\dfrac{du}{dx}-4u=2e^x\)
A1
Can be implied by next A1
\(I=\exp\int-4\,dx=e^{-4x}\)
A1ft
Must have form \(\dfrac{du}{dx}+f(x)u=g(x)\) for this mark and any further marks. Can be implied by subsequent work
\(e^{-4x}\dfrac{du}{dx}-4e^{-4x}u=2e^{-3x}\)
M1*
Multiplies through by IF of form \(e^{kx}\), simplifying RHS
\(ue^{-4x}=-\dfrac{2}{3}e^{-3x}(+A)\)
\*M1dep\*
Integrates
\(u=-\dfrac{2}{3}e^x+Ae^{4x}\)
M1dep*
Rearranges to make \(u\) or \(y^2\) the subject. No more than 1 numerical error at this step
\(y=\sqrt{-\dfrac{2}{3}e^x+Ae^{4x}}\)
A1
Cao. Ignore use of '\(\pm\)'
Alternative from 4th mark to 6th mark:
Answer
Marks
Guidance
Answer
Marks
Guidance
CF: \((u=\ldots)Ae^{4x}\)
A1
PI: \(u=ke^x\), \(\dfrac{du}{dx}=ke^x\)
M1*
PI chosen and differentiated correctly
\(ke^x-4ke^x=2e^x\), \(k=-\dfrac{2}{3}\)
M1dep*
Substitutes and solves
[8]
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u=y^2\Rightarrow\dfrac{du}{dx}=2y\dfrac{dy}{dx}$ | M1 | Correctly finds. Or $\dfrac{dy}{dx}=\dfrac{1}{2}u^{-\frac{1}{2}}\dfrac{du}{dx}$ |
| So DE $\Rightarrow 2y\dfrac{dy}{dx}-4y^2=2e^x$ | M1 | Or for complete unsimplified substitution |
| $\Rightarrow\dfrac{du}{dx}-4u=2e^x$ | A1 | Can be implied by next A1 |
| $I=\exp\int-4\,dx=e^{-4x}$ | A1ft | Must have form $\dfrac{du}{dx}+f(x)u=g(x)$ for this mark and any further marks. Can be implied by subsequent work |
| $e^{-4x}\dfrac{du}{dx}-4e^{-4x}u=2e^{-3x}$ | M1* | Multiplies through by IF of form $e^{kx}$, simplifying RHS |
| $ue^{-4x}=-\dfrac{2}{3}e^{-3x}(+A)$ | \*M1dep\* | Integrates |
| $u=-\dfrac{2}{3}e^x+Ae^{4x}$ | M1dep* | Rearranges to make $u$ or $y^2$ the subject. No more than 1 numerical error at this step |
| $y=\sqrt{-\dfrac{2}{3}e^x+Ae^{4x}}$ | A1 | Cao. Ignore use of '$\pm$' |
**Alternative from 4th mark to 6th mark:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| CF: $(u=\ldots)Ae^{4x}$ | A1 | |
| PI: $u=ke^x$, $\dfrac{du}{dx}=ke^x$ | M1* | PI chosen and differentiated correctly |
| $ke^x-4ke^x=2e^x$, $k=-\dfrac{2}{3}$ | M1dep* | Substitutes and solves |
| **[8]** | | |
---
2 Use the substitution $u = y ^ { 2 }$ to find the general solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = \frac { \mathrm { e } ^ { x } } { y }$$
for $y$ in terms of $x$.
\hfill \mbox{\textit{OCR FP3 2014 Q2 [8]}}