| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Homogeneous equation (y = vx substitution) |
| Difficulty | Standard +0.8 This is a Further Maths homogeneous differential equation requiring the substitution y=xz, separation of variables, and integration involving partial fractions (decomposing z/(1-2z²)). While the substitution is guided and the method is standard for FP3, it requires multiple sophisticated steps and careful algebraic manipulation to reach the given form, placing it moderately above average difficulty. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y=xz)\Rightarrow\dfrac{dy}{dx}=x\dfrac{dz}{dx}+z\) | B1 | For a correct statement |
| \(x\dfrac{dz}{dx}+z=\dfrac{x^2(1-z^2)}{x^2z}=\dfrac{1}{z}-z\) | M1 | For substituting into differential equation and attempting to simplify to a variables separable form |
| \(x\dfrac{dz}{dx}=\dfrac{1}{z}-2z=\dfrac{1-2z^2}{z}\) | A1 3 | For correct equation AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\dfrac{z}{1-2z^2}dz=\int\dfrac{1}{x}dx\Rightarrow-\frac{1}{4}\ln(1-2z^2)=\ln cx\) | M1, M1*, A1 | For separating variables and writing integrals; for integrating both sides to ln forms; for correct result (\(c\) not required here) |
| \(1-2z^2=(cx)^{-4}\) | A1\(\sqrt{}\) | For exponentiating their ln equation including a constant (this may follow the next M1) |
| \(\dfrac{x^2-2y^2}{x^2}=\dfrac{c^{-4}}{x^4}\) | M1(dep*) | For substituting \(z=\dfrac{y}{x}\) |
| \(x^2(x^2-2y^2)=k\) | A1 6 | For correct solution properly obtained, including dealing with any necessary change of constant to \(k\) as given AG |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y=xz)\Rightarrow\dfrac{dy}{dx}=x\dfrac{dz}{dx}+z$ | B1 | For a correct statement |
| $x\dfrac{dz}{dx}+z=\dfrac{x^2(1-z^2)}{x^2z}=\dfrac{1}{z}-z$ | M1 | For substituting into differential equation and attempting to simplify to a variables separable form |
| $x\dfrac{dz}{dx}=\dfrac{1}{z}-2z=\dfrac{1-2z^2}{z}$ | A1 **3** | For correct equation AG |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\dfrac{z}{1-2z^2}dz=\int\dfrac{1}{x}dx\Rightarrow-\frac{1}{4}\ln(1-2z^2)=\ln cx$ | M1, M1*, A1 | For separating variables and writing integrals; for integrating both sides to ln forms; for correct result ($c$ not required here) |
| $1-2z^2=(cx)^{-4}$ | A1$\sqrt{}$ | For exponentiating their ln equation including a constant (this may follow the next M1) |
| $\dfrac{x^2-2y^2}{x^2}=\dfrac{c^{-4}}{x^4}$ | M1(dep*) | For substituting $z=\dfrac{y}{x}$ |
| $x^2(x^2-2y^2)=k$ | A1 **6** | For correct solution properly obtained, including dealing with any necessary change of constant to $k$ as given AG |
---4 The variables $x$ and $y$ are related by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } - y ^ { 2 } } { x y }$$
(i) Use the substitution $y = x z$, where $z$ is a function of $x$, to obtain the differential equation
$$x \frac { \mathrm {~d} z } { \mathrm {~d} x } = \frac { 1 - 2 z ^ { 2 } } { z }$$
(ii) Hence show by integration that the general solution of the differential equation (A) may be expressed in the form $x ^ { 2 } \left( x ^ { 2 } - 2 y ^ { 2 } \right) = k$, where $k$ is a constant.
\hfill \mbox{\textit{OCR FP3 2007 Q4 [9]}}