5 Find the particular solution of the differential equation
$$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$
given that, when \(t = 0 , x = 12\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6\).
Show mark scheme
Show mark scheme source
Question 5:
Answer Marks
Guidance
Answer Marks
Guidance
\(6m^2-5m+1=0 \Rightarrow m=\frac{1}{2},\frac{1}{3}\) M1
Auxiliary equation, two distinct real roots
\(x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}\) A1
Complementary function
\(x=pt^2+qt+r \Rightarrow x'=2pt+q \Rightarrow x''=2p\) B1
Particular integral and its derivatives
\(p=1 \quad -10p+q=1 \quad 12p-5q+r=1\) M1
Substitutes and equates coefficients
\(p=1 \quad q=11 \quad r=44\) A1
\(x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}+t^2+11t+44\) A1
General solution. Need to see '\(x=\)'. FT on CF
\(x'=\frac{1}{2}Ae^{\frac{1}{2}t}+\frac{1}{3}Be^{\frac{1}{3}t}+2t+11\) M1
Differentiates
\(A+B+44=12\quad \frac{1}{2}A+\frac{1}{3}B+11=-6 \Rightarrow A=-38,\ B=6\) M1A1
Uses initial conditions
\(x = -38e^{\frac{1}{2}t}+6e^{\frac{1}{3}t}+t^2+11t+44\) A1
Need to see '\(x=\)'
10
Copy
# Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6m^2-5m+1=0 \Rightarrow m=\frac{1}{2},\frac{1}{3}$ | M1 | Auxiliary equation, two distinct real roots |
| $x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}$ | A1 | Complementary function |
| $x=pt^2+qt+r \Rightarrow x'=2pt+q \Rightarrow x''=2p$ | B1 | Particular integral and its derivatives |
| $p=1 \quad -10p+q=1 \quad 12p-5q+r=1$ | M1 | Substitutes and equates coefficients |
| $p=1 \quad q=11 \quad r=44$ | A1 | |
| $x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}+t^2+11t+44$ | A1 | General solution. Need to see '$x=$'. FT on CF |
| $x'=\frac{1}{2}Ae^{\frac{1}{2}t}+\frac{1}{3}Be^{\frac{1}{3}t}+2t+11$ | M1 | Differentiates |
| $A+B+44=12\quad \frac{1}{2}A+\frac{1}{3}B+11=-6 \Rightarrow A=-38,\ B=6$ | M1A1 | Uses initial conditions |
| $x = -38e^{\frac{1}{2}t}+6e^{\frac{1}{3}t}+t^2+11t+44$ | A1 | Need to see '$x=$' |
| | **10** | |
---
Show LaTeX source
Copy
5 Find the particular solution of the differential equation
$$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$
given that, when $t = 0 , x = 12$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6$.\\[0pt]
[10]\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-10_2715_40_110_2007}\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-11_2726_35_97_20}\\
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q5 [10]}}