Question 5:
5 | (a) | 1 dv dv 1 1
F =ma⇒ t−v=2 ⇒ + v= t
2 dt dt 2 4 | B1
[1] | 3.3 | AG. Intermediate step as shown
or both sides halved
(b) | IF =
1 1
∫2𝑑𝑑𝑑𝑑 2𝑑𝑑
1 t dv𝑒𝑒 1 1 =t 𝑒𝑒 d  1 t  1 1 t
e2 + e2 v= ve2 = te2
dt 2 dt  4
1 1 1 1
2 𝑑𝑑 1 2 𝑑𝑑 1 2 𝑑𝑑 1 2 𝑑𝑑
𝑣𝑣𝑒𝑒 = 𝑡𝑡𝑒𝑒 − �𝑒𝑒 𝑎𝑎𝑡𝑡 = 𝑡𝑡𝑒𝑒
2 2 2
0=−1+c⇒c=1 1 𝑑𝑑
2
−𝑒𝑒 (+𝑐𝑐)
1 − 1 t
v= t−1+e 2
2 | B1
M1
A1
M1
A1 | 1.1
1.1
1.1
3.4
1.1 | 1 1
t+c t
Or e2 or Ae2
Multiplying both sides by IF and
writing new LHS as an exact
derivative
Use initial conditions to
determine c | Allow if using a solution of
correct form with wrong (non-
zero) coefficients
Alternate method
AE:
1 1
CF: 𝜆𝜆+2 = 0,𝜆𝜆 = −2 | B1
1
Trial function−: 2𝑑𝑑
( )
𝑣𝑣 = 𝐴𝐴e
𝑑𝑑𝑑𝑑
𝑣𝑣 = 𝑎𝑎 𝑡𝑡+𝑏𝑏, 𝑑𝑑𝑑𝑑 = 𝑎𝑎
1 1 | M1A1
𝑎𝑎+ (𝑎𝑎𝑡𝑡+𝑏𝑏)= 𝑡𝑡
2 4
1
𝑎𝑎GS=: ,𝑏𝑏 = −1
2 1
−2𝑑𝑑 1 | M1 | Substitutes boundary condition
into general solution to find A | Allow if using a solution of
correct form with wrong (non-
zero) coefficients
(𝑣𝑣 =)𝐴𝐴e +2𝑡𝑡−1
𝑣𝑣 = 0,𝑡𝑡 = 0 gives 𝐴𝐴 = 1
1
− 2 𝑑𝑑 1
M1A1
M1
Substitutes boundary condition
into general solution to find A
Allow if using a solution of
correct form with wrong (non-
zero) coefficients
A1
[5]
(c) | t = 2 => v=e−1 (so velocity is e–1 ms-1) | B1
[1] | 3.4 | or awrt 0.368
(d) | change the RHS of DE in part (a) from 1tto
4
1 . oe
4 | B1
[1] | 3.5c | .
(e) | d  1 t  1 1 t 1 t 1 1 t
ve2 = e2 ⇒ve2 =∫ e2 dt
dt  4 4
1 t 1 1 t
ve2 = e2 +c. oe
2
1 1
e−1e1 = e1+c⇒c=1− e
2 2
1  1  − 1 t
v= +1− ee 2
2  2  | B1
M1
A1
[3] | 2.2a
3.3
1.1 | SC1 oe
1 1
2𝑑𝑑 𝑁𝑁 2𝑑𝑑
𝑣𝑣𝑒𝑒 = 2𝑒𝑒 +𝑐𝑐
substitute their boundary
condition from (c) into correct
GS to find c | dv 1 1
Or + v= rearranged to
dt 2 4
giving
2𝑑𝑑𝑑𝑑 1 1
1
-∫ l1n−(21𝑑𝑑−=2v2)∫+𝑎𝑎c𝑡𝑡'== 2t𝑡𝑡
2
′ −1
𝑐𝑐 = 1+ln (1−2𝑒𝑒 )
(b) | So x = ±0.1cos(5πt) or x=0.1sin(5πt± 1π)
2
when t = 0.75 or 0.35
2
x=− (=−0.0707to 3 sf)
20 | M1
A1
[2] | 3.1b
3.4 | or by argument from sketch | Condone amplitude of 0.2 for M1