MATHEMATICS 2 (MA10193) EXAMPLES SHEET 4: SOLUTIONS

1. The auxiliary equation is λ^{2}−8λ+41 = 0, which has solutions λ = 4±5i. So the general solution is y = Ae^{4x} cos 5x + B4x sin 5x. (We could write this in terms of exponentials if we preferred.) The boundary condition at x = 0 gives A = 1 and the boundary condition at x = 5π/2 gives (remember that cos 5π/2 = 0, sin 5π/2 = 1) gives B = 0. So the solution is y = e^{4x} cos 5x.

2. The auxiliary equation is λ^{2} − 3λ + 10 = 0, which has solutions λ = (3 ± i\sqrt{31})/2. A particular integral is of the form y = α sin x + β cos x for suitable α and β, and the condition is

−α sin x − β cos x − 3α cos x + 3β sin x + 10α sin x + 10β cos x = 24 sin x,

i.e. (9α + 3β) sin x + (−3α + 9β) cos x = 24 sin x. Therefore

\begin{aligned}9α + 3β & = 24 \\ −3α + 9β & = 0 \end{aligned}

which gives α = 12/5, β = 4/5. So the general solution is

y = Ae^{3x/2} cos \sqrt{31}x/2 + Be^{3x/2} sin \sqrt{31}x/2 + \frac{12}{5} sin x + \frac{4}{5} cos x.

The boundary condition at x = 0 gives

0 = A + \frac{4}{5}

so A = −4/5, and the condition at x = π/4 gives

1 = −\frac{4}{5}e^{3π/4} cos \sqrt{31}π/4 + Be^{3π/4} sin \sqrt{31}π/4 + \frac{8\sqrt{2}}{5},

since cos π/2 = sin π/2 = \sqrt{2}/2. Therefore

B = (1 + \frac{4}{5}e^{3π/4} cos \sqrt{31}π/4 − \frac{8\sqrt{2}}{5})/(e^{3π/4} sin \sqrt{31}π/4) ≈ 0.409.

3. From the first equation, 3y = −7\frac{dx}{dt} + x. Putting this into the second equation gives

−7\frac{d^{x}}{dt^{2}} + \frac{dx}{dt} − 4x = 0.

The auxiliary equation for this is −7λ^{2} + λ − 4 = 0, so λ = (1 ± i\sqrt{111})/14. Call these λ_{1} and λ_{2}. Then x = Ae^{λ_{1}t} + Be^{λ_{2}t} and therefore y = \frac{−7λ_{1}+1}{3}Ae^{λ_{1}t} + \frac{−7λ_{2}+1}{3}Be^{λ_{2}t}.

4. The best way to do this is to eliminate \frac{dx}{dt}. If we multiply the first equation by 2 and subtract the first equation from it we get

5\frac{dy}{dt} + 10x − 13y = 6t^{2} + 2t − 2,

so

x = −\frac{1}{2} \frac{dy}{dt} + \frac{13}{10}y + \frac{3}{5}t^{2} + \frac{1}{5}t − \frac{1}{5}.

Substituting this in the second equation we get

−\frac{d^{y}}{dt^{2}} + \frac{13}{5} \frac{dy}{dt} − \frac{12}{5}t + \frac{2}{5} + \frac{dy}{dt} − 2\frac{dy}{dt} + \frac{26}{5}y + \frac{12}{5}t^{2} + \frac{4}{5}t − \frac{4}{5} + 5y = −2t + 2,

that is (collecting the terms and multiplying by 5)

5\frac{d^{y}}{dt^{2}} − 8\frac{dy}{dt} − 51y = 60t^{2} + 130t − 60.

The CF for this is Ae^{λ_{1}t} + Be^{λ_{2}t} where λ_{1} = (4 + \sqrt{271})/5, λ_{2} = (4 − \sqrt{271})/5 A PI is of the form at^{2} + bt + c and we get

60t^{2} + 130t − 60 = 10a − 16at − 8b − 51at^{2} − 51bt − 51c

so a = −60/51, b = −630/289 and c = 18980/14739. Hence y = Ae^{(4+\sqrt{271})t/5} + Be^{(4−\sqrt{271})t/5} − 60t^{2}/51 − 630t/289 + 18980/14739, and x can be calculated from the formula above: in floating point

x = −2.79Ae^{4.09t} + 3.79Be^{−2.49t} − 0.93t^{2} − 0.28t + 3.65.

GKS, 9/05/05