CAIE FP1 2013 June — Question 9

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations

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 \(\lim _ { t \rightarrow \infty } x\).

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}}