9 Find \(x\) in terms of \(t\) given that
$$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 6 \mathrm { e } ^ { - 2 t }$$
and that, when \(t = 0 , x = \frac { 5 } { 3 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 7 } { 6 }\).
State \(\lim _ { t \rightarrow \infty } x\).
Show mark scheme
Show mark scheme source
Question 9:
Finds complementary function:
Answer Marks
\(4m^2 + 4m + 1 = 0 \Rightarrow (2m+1)^2 = 0 \Rightarrow m = -\frac{1}{2}\) M1
C.F.: \(x = Ae^{-\frac{t}{2}} + Bte^{-\frac{t}{2}}\) A1
Finds particular integral:
Answer Marks
P.I.: \(x = ke^{-2t} \Rightarrow \dot{x} = -2ke^{-2t} \Rightarrow \ddot{x} = 4ke^{-2t}\) M1
\(16k - 8k + k = 6 \Rightarrow k = \frac{2}{3} \Rightarrow x = \frac{2}{3}e^{-2t}\) A1
Adds for general solution:
Answer Marks
G.S.: \(x = Ae^{-\frac{t}{2}} + Bte^{-\frac{t}{2}} + \frac{2}{3}e^{-2t}\) A1
Uses initial conditions to find constants:
Answer Marks
\(x(0) = \frac{5}{3} \Rightarrow \frac{5}{3} = A + \frac{2}{3} \Rightarrow A = 1\) B1
\(\dot{x} = -\frac{1}{2}e^{-\frac{t}{2}} + Be^{-\frac{t}{2}} - \frac{1}{2}Bte^{-\frac{t}{2}} - \frac{4}{3}e^{-2t}\) M1
\(\dot{x}(0) = \frac{7}{6} \Rightarrow \frac{7}{6} = -\frac{1}{2} + B - \frac{4}{3} \Rightarrow B = 3\) A1
Gives particular solution:
Answer Marks
Guidance
\(x = e^{-\frac{t}{2}} + 3te^{-\frac{t}{2}} + \frac{2}{3}e^{-2t}\) A1
9 marks
States limit:
Answer Marks
Guidance
\(\lim_{t \to \infty} x = 0\) B1
1 mark; [10 total]
Copy
## Question 9:
**Finds complementary function:**
| $4m^2 + 4m + 1 = 0 \Rightarrow (2m+1)^2 = 0 \Rightarrow m = -\frac{1}{2}$ | M1 | |
| C.F.: $x = Ae^{-\frac{t}{2}} + Bte^{-\frac{t}{2}}$ | A1 | |
**Finds particular integral:**
| P.I.: $x = ke^{-2t} \Rightarrow \dot{x} = -2ke^{-2t} \Rightarrow \ddot{x} = 4ke^{-2t}$ | M1 | |
| $16k - 8k + k = 6 \Rightarrow k = \frac{2}{3} \Rightarrow x = \frac{2}{3}e^{-2t}$ | A1 | |
**Adds for general solution:**
| G.S.: $x = Ae^{-\frac{t}{2}} + Bte^{-\frac{t}{2}} + \frac{2}{3}e^{-2t}$ | A1 | |
**Uses initial conditions to find constants:**
| $x(0) = \frac{5}{3} \Rightarrow \frac{5}{3} = A + \frac{2}{3} \Rightarrow A = 1$ | B1 | |
| $\dot{x} = -\frac{1}{2}e^{-\frac{t}{2}} + Be^{-\frac{t}{2}} - \frac{1}{2}Bte^{-\frac{t}{2}} - \frac{4}{3}e^{-2t}$ | M1 | |
| $\dot{x}(0) = \frac{7}{6} \Rightarrow \frac{7}{6} = -\frac{1}{2} + B - \frac{4}{3} \Rightarrow B = 3$ | A1 | |
**Gives particular solution:**
| $x = e^{-\frac{t}{2}} + 3te^{-\frac{t}{2}} + \frac{2}{3}e^{-2t}$ | A1 | 9 marks |
**States limit:**
| $\lim_{t \to \infty} x = 0$ | B1 | 1 mark; **[10 total]** |
---
Show LaTeX source
Copy
9 Find $x$ in terms of $t$ given that
$$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 6 \mathrm { e } ^ { - 2 t }$$
and that, when $t = 0 , x = \frac { 5 } { 3 }$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 7 } { 6 }$.
State $\lim _ { t \rightarrow \infty } x$.
\hfill \mbox{\textit{CAIE FP1 2013 Q9 [10]}}