OCR FP3 2013 June — Question 3 8 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeSubstitution reducing to first order linear ODE
DifficultyChallenging +1.2 This is a Further Maths FP3 question requiring recognition that a substitution simplifies to a first-order linear ODE solvable by integrating factor. The substitution is given, making it more routine than if students had to identify it themselves. The mechanics involve differentiating u = y³, substituting to get a linear form, finding an integrating factor (x³), and integrating. While this requires multiple steps and is from Further Maths content, the substitution being provided and the standard integrating factor technique make it moderately above average difficulty rather than highly challenging.
Spec4.10a General/particular solutions: of differential equations
3 The differential equation $$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y ^ { 3 } = \frac { \cos x } { x }$$ is to be solved for \(x > 0\). Use the substitution \(u = y ^ { 3 }\) to find the general solution for \(y\) in terms of \(x\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(u = y^3 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\)M1 Or \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3}u^{-\frac{2}{3}}\frac{\mathrm{d}u}{\mathrm{d}x}\)
In DE gives \(x\frac{\mathrm{d}u}{\mathrm{d}x} + 2u = \frac{\cos x}{x}\)A1
\(\frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2}{x}u = \frac{\cos x}{x^2}\)B1 Divide both sides. Must have form \(\frac{\mathrm{d}u}{\mathrm{d}x} + f(x)u = g(x)\)
\(I = \exp\!\left(\int \frac{2}{x}\,\mathrm{d}x\right) = e^{2\ln x}\)M1 Correctly integrates. Can be implied by subsequent work
\(= x^2\)A1
\(x^2\frac{\mathrm{d}u}{\mathrm{d}x} + 2xu = \cos x\)
\(\frac{\mathrm{d}}{\mathrm{d}x}(x^2 u) = \cos x\)
\(x^2 u = \sin x \quad (+A)\)M1 Integrate
\(u = \frac{\sin x + A}{x^2}\)A1 Or gives GS in implicit form. Must include constant at this stage
\(y = \left(\frac{\sin x + A}{x^2}\right)^{\frac{1}{3}}\)A1 
# Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = y^3 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 3y^2\frac{\mathrm{d}y}{\mathrm{d}x}$ | M1 | Or $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3}u^{-\frac{2}{3}}\frac{\mathrm{d}u}{\mathrm{d}x}$ |
| In DE gives $x\frac{\mathrm{d}u}{\mathrm{d}x} + 2u = \frac{\cos x}{x}$ | A1 | |
| $\frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2}{x}u = \frac{\cos x}{x^2}$ | B1 | Divide both sides. Must have form $\frac{\mathrm{d}u}{\mathrm{d}x} + f(x)u = g(x)$ |
| $I = \exp\!\left(\int \frac{2}{x}\,\mathrm{d}x\right) = e^{2\ln x}$ | M1 | Correctly integrates. Can be implied by subsequent work |
| $= x^2$ | A1 | |
| $x^2\frac{\mathrm{d}u}{\mathrm{d}x} + 2xu = \cos x$ | | |
| $\frac{\mathrm{d}}{\mathrm{d}x}(x^2 u) = \cos x$ | | |
| $x^2 u = \sin x \quad (+A)$ | M1 | Integrate |
| $u = \frac{\sin x + A}{x^2}$ | A1 | Or gives GS in implicit form. Must include constant at this stage |
| $y = \left(\frac{\sin x + A}{x^2}\right)^{\frac{1}{3}}$ | A1 | |

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3 The differential equation

$$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y ^ { 3 } = \frac { \cos x } { x }$$

is to be solved for $x > 0$. Use the substitution $u = y ^ { 3 }$ to find the general solution for $y$ in terms of $x$.

\hfill \mbox{\textit{OCR FP3 2013 Q3 [8]}}
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