7. (a) Show that the transformation \(x = t ^ { 2 }\) transforms the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 15 y = 15 x$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 15 y = 15 t ^ { 2 }$$
(b) Solve differential equation (II) to determine \(y\) in terms of \(t\).
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Question 7:
Part (a):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(x = t^2 \Rightarrow \frac{dx}{dy} = 2t\frac{dt}{dy}\) oe M1
Correct application of the chain rule
\(\Rightarrow \frac{dy}{dx} = \frac{1}{2t}\frac{dy}{dt}\) (or e.g. \(\frac{1}{2\sqrt{x}}\frac{dy}{dt}\)) A1
Any correct expression for \(\frac{dy}{dx}\) or equivalent equation
\(2\sqrt{x}\frac{dy}{dx} = \frac{dy}{dt} \Rightarrow x^{-\frac{1}{2}}\frac{dy}{dx} + 2\sqrt{x}\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}\frac{dt}{dx}\) (NB \(\frac{d^2y}{dt^2} = 2\frac{dy}{dx} + 4x\frac{d^2y}{dx^2}\)) dM1
Fully correct strategy to obtain equation involving \(\frac{d^2y}{dx^2}\) and \(\frac{d^2y}{dt^2}\). Chain rule used on at least one term. Depends on first M mark
\(4x\frac{d^2y}{dx^2} + 2(1+2\sqrt{x})\frac{dy}{dx} - 15y = 15x \Rightarrow 4x\frac{d^2y}{dx^2} + 4\sqrt{x}\frac{dy}{dx} + 2\frac{dy}{dx} - 15y = 15x\) \(\Rightarrow \frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 15y = 15t^2\)* ddM1 A1*
ddM1: Substitutes into the given differential equation. Full substitution must be seen. Depends on both M marks. A1*: Cso
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(m^2 + 2m - 15 = 0 \Rightarrow m = 3, -5\) M1
Attempts to solve \(m^2 + 2m - 15 = 0\)
\(y = Ae^{-5t} + Be^{3t}\) A1
Correct CF
\(y = at^2 + bt + c \Rightarrow \frac{dy}{dt} = 2at + b \Rightarrow \frac{d^2y}{dt^2} = 2a\) \(\Rightarrow 2a + 4at + 2b - 15at^2 - 15bt - 15c = 15t^2\) M1
Starts with the correct PI form and differentiates twice and substitutes
\(-15a = 15 \Rightarrow a = \ldots\) \(4a - 15b = 0 \Rightarrow b = \ldots\) \(2a + 2b - 15c = 0 \Rightarrow c = \ldots\) dM1
Complete method to find \(a\), \(b\) and \(c\) by comparing coefficients. Values for all 3 needed. Depends on second M mark
\(y = Ae^{-5t} + Be^{3t} - t^2 - \frac{4}{15}t - \frac{38}{225}\) A1
Correct GS. Must start \(y = \ldots\)
Part (c):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(y = Ae^{-5\sqrt{x}} + Be^{3\sqrt{x}} - x - \frac{4}{15}\sqrt{x} - \frac{38}{225}\) B1ft
Correct equation (follow through their answer to (b)). Must start \(y = \ldots\)
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## Question 7:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = t^2 \Rightarrow \frac{dx}{dy} = 2t\frac{dt}{dy}$ oe | M1 | Correct application of the chain rule |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{2t}\frac{dy}{dt}$ (or e.g. $\frac{1}{2\sqrt{x}}\frac{dy}{dt}$) | A1 | Any correct expression for $\frac{dy}{dx}$ or equivalent equation |
| $2\sqrt{x}\frac{dy}{dx} = \frac{dy}{dt} \Rightarrow x^{-\frac{1}{2}}\frac{dy}{dx} + 2\sqrt{x}\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}\frac{dt}{dx}$ (NB $\frac{d^2y}{dt^2} = 2\frac{dy}{dx} + 4x\frac{d^2y}{dx^2}$) | dM1 | Fully correct strategy to obtain equation involving $\frac{d^2y}{dx^2}$ and $\frac{d^2y}{dt^2}$. Chain rule used on at least one term. Depends on first M mark |
| $4x\frac{d^2y}{dx^2} + 2(1+2\sqrt{x})\frac{dy}{dx} - 15y = 15x \Rightarrow 4x\frac{d^2y}{dx^2} + 4\sqrt{x}\frac{dy}{dx} + 2\frac{dy}{dx} - 15y = 15x$ $\Rightarrow \frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 15y = 15t^2$* | ddM1 A1* | ddM1: Substitutes into the given differential equation. Full substitution must be seen. Depends on both M marks. A1*: Cso |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 2m - 15 = 0 \Rightarrow m = 3, -5$ | M1 | Attempts to solve $m^2 + 2m - 15 = 0$ |
| $y = Ae^{-5t} + Be^{3t}$ | A1 | Correct CF |
| $y = at^2 + bt + c \Rightarrow \frac{dy}{dt} = 2at + b \Rightarrow \frac{d^2y}{dt^2} = 2a$ $\Rightarrow 2a + 4at + 2b - 15at^2 - 15bt - 15c = 15t^2$ | M1 | Starts with the correct PI form and differentiates twice and substitutes |
| $-15a = 15 \Rightarrow a = \ldots$ $4a - 15b = 0 \Rightarrow b = \ldots$ $2a + 2b - 15c = 0 \Rightarrow c = \ldots$ | dM1 | Complete method to find $a$, $b$ and $c$ by comparing coefficients. Values for all 3 needed. Depends on second M mark |
| $y = Ae^{-5t} + Be^{3t} - t^2 - \frac{4}{15}t - \frac{38}{225}$ | A1 | Correct GS. Must start $y = \ldots$ |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = Ae^{-5\sqrt{x}} + Be^{3\sqrt{x}} - x - \frac{4}{15}\sqrt{x} - \frac{38}{225}$ | B1ft | Correct equation (follow through their answer to (b)). Must start $y = \ldots$ |
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7. (a) Show that the transformation $x = t ^ { 2 }$ transforms the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 15 y = 15 x$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 15 y = 15 t ^ { 2 }$$
(b) Solve differential equation (II) to determine $y$ in terms of $t$.\\
(c) Hence determine the general solution of differential equation (I).\\
\includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-24_2258_53_308_1980}
\hfill \mbox{\textit{Edexcel F2 2021 Q7 [11]}}