Challenging +1.2 This is a standard Euler-Cauchy equation problem from FP2 following a well-established template. Part (a) is routine verification of a given substitution using chain rule, part (b) is solving a constant-coefficient second-order DE (standard auxiliary equation method), and part (c) is back-substitution. While it requires multiple techniques and careful algebra, it's a textbook exercise with no novel insight required, making it moderately above average difficulty for A-level but standard for Further Maths.
8. (a) Show that the substitution \(x = \mathrm { e } ^ { z }\) transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 3 \ln x , \quad x > 0$$
into the equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} z ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} z } - 2 y = 3 z$$
(b) Find the general solution of the differential equation (II).
(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form \(y = \mathrm { f } ( x )\).
\(\square\)
\(y = Ax^{-2} + Bx - \frac{3}{2}\ln x - \frac{3}{4}\)
B1ft
Reverses substitution to obtain solution in the form \(y = \ldots\); follow through from complete solution in (b)
[14]
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## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = e^z$ | | |
| $\frac{dx}{dy} = e^z\frac{dz}{dy}$ | M1 | Differentiates $x=e^z$ wrt $y$; chain rule must be used |
| $\frac{dy}{dx} = e^{-z}\frac{dy}{dz}$ | A1 | Correct differentiation |
| $\frac{d^2y}{dx^2} = -e^{-z}\frac{dz}{dx}\times\frac{dy}{dz} + e^{-z}\frac{d^2y}{dz^2}\times\frac{dz}{dx} = \frac{1}{x^2}\left(-\frac{dy}{dz}+\frac{d^2y}{dz^2}\right)$ | M1A1A1 | Differentiates again to obtain $\frac{d^2y}{dx^2}$; one mark for each correct term |
| $x^2\left(-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x^2}\frac{d^2y}{dz^2}\right) + 2x\times\frac{1}{x}\frac{dy}{dz} - 2y = 3z$ | M1 | Substitutes into the given equation |
| $\dfrac{d^2y}{dz^2} + \dfrac{dy}{dz} - 2y = 3z$ | A1 | cso; obtains the required equation | (7) |
**Alt** ($z = \ln x$):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dz}\times\frac{dz}{dx} = \frac{1}{x}\frac{dy}{dz}$ | M1A1 | |
| $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dz} + \frac{1}{x}\frac{d^2y}{dz^2}\times\frac{dz}{dx} = -\frac{1}{x^2}\frac{dy}{dz} + \frac{1}{x^2}\frac{d^2y}{dz^2}$ | M1A1A1 | |
| Substitution and simplification gives $\dfrac{d^2y}{dz^2} + \dfrac{dy}{dz} - 2y = 3z$ | M1A1 | (7) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Auxiliary equation: $m^2 + m - 2 = 0$; $(m+2)(m-1)=0$; $m = -2, 1$ | M1A1 | Forms and solves auxiliary equation; both values correct |
| CF: $y = Ae^{-2z} + Be^{z}$ | A1 | Correct CF |
| PI: Try $y = az + b$; $\frac{dy}{dz}=a$, $\frac{d^2y}{dz^2}=0$ | M1 | Tries suitable expression for PI |
| $a - 2(az+b) = 3z \Rightarrow a = -\frac{3}{2},\ b = -\frac{3}{4}$ | | Obtains values for constants in PI |
| Complete solution: $y = Ae^{-2z} + Be^{z} - \frac{3}{2}z - \frac{3}{4}$ | A1A1 | One mark for each correct term in PI | (6) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = Ax^{-2} + Bx - \frac{3}{2}\ln x - \frac{3}{4}$ | B1ft | Reverses substitution to obtain solution in the form $y = \ldots$; follow through from complete solution in (b) | (1) |
| | | **[14]** |
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8. (a) Show that the substitution $x = \mathrm { e } ^ { z }$ transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 3 \ln x , \quad x > 0$$
into the equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} z ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} z } - 2 y = 3 z$$
(b) Find the general solution of the differential equation (II).\\
(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form $y = \mathrm { f } ( x )$.\\
\(\square\)
\hfill \mbox{\textit{Edexcel FP2 2014 Q8 [14]}}