Standard +0.8 This is a second-order vector differential equation requiring auxiliary equation solution, application of initial conditions in vector form, and evaluation at a specific logarithmic time value. While the method is standard for Further Maths, the vector nature adds complexity and the ln 2 evaluation requires careful algebraic manipulation, placing it moderately above average difficulty.
3. At time \(t\) seconds, the position vector \(\mathbf { r }\) metres of a particle \(P\), relative to a fixed origin \(O\), satisfies the differential equation
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 3 \mathbf { r } = \mathbf { 0 }$$
At time \(t = 0 , P\) is at the point with position vector \(2 \mathbf { i } \mathrm {~m}\) and is moving with velocity \(2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Find the position vector of \(P\) when \(t = \ln 2\).
(10 marks)
3. At time $t$ seconds, the position vector $\mathbf { r }$ metres of a particle $P$, relative to a fixed origin $O$, satisfies the differential equation
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 3 \mathbf { r } = \mathbf { 0 }$$
At time $t = 0 , P$ is at the point with position vector $2 \mathbf { i } \mathrm {~m}$ and is moving with velocity $2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Find the position vector of $P$ when $t = \ln 2$.\\
(10 marks)\\
\hfill \mbox{\textit{Edexcel M5 Q3 [10]}}