5 Solve the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = \mathrm { e } ^ { - x }$$
subject to the conditions \(y = \frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).
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Question 5:
Answer Marks
Guidance
Answer Marks
Guidance
AE: \(\lambda^2 + 5\lambda + 6 = 0\); \(\lambda = -2, -3\) B1
CF: \(Ae^{-2x} + Be^{-3x}\) B1ft
PI: \(y = ae^{-x}\) B1ft
\(ae^{-x} - 5ae^{-x} + 6ae^{-x} = e^{-x}\); \(2a = 1\) M1
Differentiate and substitute
\(a = \frac{1}{2}\) A1
GS: \(y = \frac{1}{2}e^{-x} + Ae^{-2x} + Be^{-3x}\) A1ft
ft must be of form "\(ke^{-x}\) plus a standard CF form" with 2 arbitrary constants
\(x=0, y=0 \Rightarrow \frac{1}{2} + A + B = 0\) M1
Use condition on GS; Must have 2 arbitrary constants
\(y' = -\frac{1}{2}e^{-x} - 2Ae^{-2x} - 3Be^{-3x}\) M1*
Differentiate their GS of form \(y = ke^{-x} + Ae^{mx} + Be^{nx}\) where k, m, n are non-zero constants and m, n not 1
\(x=0, y'=0 \Rightarrow -\frac{1}{2} - 2A - 3B = 0\) \(A = -1, B = \frac{1}{2}\) M1dep*
Use condition and attempt to find A, B
\(y = \frac{1}{2}e^{-x} - e^{-2x} + \frac{1}{2}e^{-3x}\) A1
www; Must have 'y ='
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# Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| AE: $\lambda^2 + 5\lambda + 6 = 0$; $\lambda = -2, -3$ | B1 | |
| CF: $Ae^{-2x} + Be^{-3x}$ | B1ft | |
| PI: $y = ae^{-x}$ | B1ft | |
| $ae^{-x} - 5ae^{-x} + 6ae^{-x} = e^{-x}$; $2a = 1$ | M1 | Differentiate and substitute |
| $a = \frac{1}{2}$ | A1 | |
| GS: $y = \frac{1}{2}e^{-x} + Ae^{-2x} + Be^{-3x}$ | A1ft | ft must be of form "$ke^{-x}$ plus a standard CF form" with 2 arbitrary constants |
| $x=0, y=0 \Rightarrow \frac{1}{2} + A + B = 0$ | M1 | Use condition on GS; Must have 2 arbitrary constants |
| $y' = -\frac{1}{2}e^{-x} - 2Ae^{-2x} - 3Be^{-3x}$ | M1* | Differentiate their GS of form $y = ke^{-x} + Ae^{mx} + Be^{nx}$ where k, m, n are non-zero constants and m, n not 1 |
| $x=0, y'=0 \Rightarrow -\frac{1}{2} - 2A - 3B = 0$ | | |
| $A = -1, B = \frac{1}{2}$ | M1dep* | Use condition and attempt to find A, B |
| $y = \frac{1}{2}e^{-x} - e^{-2x} + \frac{1}{2}e^{-3x}$ | A1 | www; Must have 'y =' |
| **[10]** | | |
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5 Solve the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = \mathrm { e } ^ { - x }$$
subject to the conditions $y = \frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = 0$.
\hfill \mbox{\textit{OCR FP3 2014 Q5 [10]}}