8. (a) Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = 0 , \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 0$$
(b) Hence find the general solution of the differential equation (I).
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Question 8:
\(x = e^t\)
Part (a):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\) M1
Attempt to use an appropriate version of the chain rule
\(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{1}{e^t}\left(= \frac{1}{x}\frac{dy}{dt}\right)\) A1
oe
\(\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\frac{d^2y}{dt^2}\frac{dt}{dx}\) or \(\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x^2}\frac{d^2y}{dt^2}\) M1A1,A1
M1: Use of the product rule (penalise chain rule errors by loss of A mark). Note \(t = \ln x \Rightarrow \frac{dt}{dx} = \frac{1}{x}\)
\(x^2\frac{d^2y}{dx^2} + 5x\frac{dy}{dx} + 13y = 0 \Rightarrow x^2 \cdot \frac{1}{x^2}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) + 5x\frac{1}{x}\frac{dy}{dt} + 13y = 0\) ddM1A1
M1: Substitutes first and second derivatives into the given DE. Depends on both M marks above. A1: Correct completion to printed answer
\(\Rightarrow \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 13y = 0^*\)
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(m^2 + 4m + 13 = 0 \Rightarrow (m=)\frac{-4\pm\sqrt{16-52}}{2}\) M1
Attempt to solve the auxiliary equation
\((m=) -2 \pm 3i\) A1
Correct roots. May be implied by a correct GS
\(y = e^{-2t}(A\cos 3t + B\sin 3t)\) or \(y = Ae^{(-2+3i)t} + Be^{(-2-3i)t}\) A1
Correct GS
\(t = \ln x\) B1
\(y = \frac{A\cos(3\ln x) + B\sin(3\ln x)}{x^2}\) or \(y = Ae^{(-2+3i)\ln x} + Be^{(-2-3i)\ln x}\) A1
Total: 12
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## Question 8:
$x = e^t$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$ | M1 | Attempt to use an appropriate version of the chain rule |
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{1}{e^t}\left(= \frac{1}{x}\frac{dy}{dt}\right)$ | A1 | oe |
| $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\frac{d^2y}{dt^2}\frac{dt}{dx}$ or $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x^2}\frac{d^2y}{dt^2}$ | M1A1,A1 | M1: Use of the product rule (penalise chain rule errors by loss of A mark). Note $t = \ln x \Rightarrow \frac{dt}{dx} = \frac{1}{x}$ |
| $x^2\frac{d^2y}{dx^2} + 5x\frac{dy}{dx} + 13y = 0 \Rightarrow x^2 \cdot \frac{1}{x^2}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) + 5x\frac{1}{x}\frac{dy}{dt} + 13y = 0$ | ddM1A1 | M1: Substitutes first and second derivatives into the given DE. Depends on both M marks above. A1: Correct completion to printed answer |
| $\Rightarrow \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 13y = 0^*$ | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 4m + 13 = 0 \Rightarrow (m=)\frac{-4\pm\sqrt{16-52}}{2}$ | M1 | Attempt to solve the auxiliary equation |
| $(m=) -2 \pm 3i$ | A1 | Correct roots. May be implied by a correct GS |
| $y = e^{-2t}(A\cos 3t + B\sin 3t)$ or $y = Ae^{(-2+3i)t} + Be^{(-2-3i)t}$ | A1 | Correct GS |
| $t = \ln x$ | B1 | |
| $y = \frac{A\cos(3\ln x) + B\sin(3\ln x)}{x^2}$ or $y = Ae^{(-2+3i)\ln x} + Be^{(-2-3i)\ln x}$ | A1 | |
**Total: 12**
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8. (a) Show that the substitution $x = \mathrm { e } ^ { t }$ transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = 0 , \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 0$$
(b) Hence find the general solution of the differential equation (I).\\
\hfill \mbox{\textit{Edexcel F2 2014 Q8 [12]}}