Challenging +1.3 This is a Further Maths question requiring elimination to form a second-order ODE, solving the auxiliary equation, applying initial conditions, and back-substituting to find both variables. While systematic, it demands multiple techniques beyond standard A-level and careful algebraic manipulation throughout, placing it notably above average difficulty.
14 Solve the simultaneous differential equations
\(\frac { \mathrm { d } x } { \mathrm {~d} t } + 2 x = 4 y , \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 x = 5 y\),
given that when \(t = 0 , x = 0\) and \(y = 1\).
Question 14:
14 | ⇒
⇒
AE λ2 − 3λ + 2 = 0
⇒λ = 1 or 2
GS x = Aet + Be2 t
when t = 0, x = A + B = 0
⇒ A = −4, B = 4
so x = 4e2 t – 4et
y = 4e2 t – 3et | M1
M1
A1
B1ft
M1
A1
M1
A1
M1
A1
A1
[11] | 3.1a
3.1a
1.1
1.1
2.1
2.2a
1.1
1.1
1.1
3.2a
3.2a | diff and subst for dy/dt or
dx/dt
subst for y (or x)
Must be simplified
ft their values of λ
subst for x, dx/dt
subst t = 0 in x, y to find eqns
in A, B
both equations correct
solving the equations to find
A and B
for both A and B
for correct equations for both
x and y
| d2y +3dx = 5 dy
dt2 dt dt
1(5 dy − d2y )+ 2(5y− dy ) = 4y
3 dt dt2 3 dt
d2y dy
−3 +2y = 0
dt2 dt
AE λ2 − 3λ + 2 = 0
⇒λ = 1 or 2
GS y = Cet + De2 t
x = 4Cet+De2t
3
when t = 0, 4C + 3D = 0
C + D = 1
C = -3 D = 4
14 Solve the simultaneous differential equations\\
$\frac { \mathrm { d } x } { \mathrm {~d} t } + 2 x = 4 y , \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 x = 5 y$,\\
given that when $t = 0 , x = 0$ and $y = 1$.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q14 [11]}}