Week05 solns - Tutorial sessions PDF

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Maths Problems for CHEN10072 (With solutions) Bill Lionheart

Week 5 1. Solve analytically the following differential equations: dy = −x2 y ; y = 1 when x = 0. dx dy = −y2 sin x ; y = 1 when x = π/2. (b) dx dy (c) = −x(4 + y2 ) ; y = 2 when x = 0. dx (a)

Solution (a)

Z

dy y

=

Z

−x2 dx

ln |y| = −

x3 +c 3

y = Ae−

x3 3

Note we removed the absolute value as A can be positive or negative. Using y = 1 when x = 0 we see A = 1 and y = e−

x3 3

is the particular solution. (b)

Z

Z dy = − sin x dx y2 1 = − cos x + c y 1 y = c − cos x

Using y = 1 when x = π/2 we see have 1 = −0 + c so c = 1 and y=

1 1 − cos x

is the particular solution. (c)

Z

dy 4 = y2 y 1 tan−1 2 2 y tan−1 2 y 1

=

Z

−x dx

x2 +c 2 x2 = − +C 2 = 2 tan(C − x2 /2) = −

Now y = 2 when x = 0 gives 2 = 2 tan C, so C = π/4 and the general solution is y = 2 tan(π/4 − x2 /2)

2. Consider the second order reaction between species A and B, A + B

−→

Products

with initial concentrations [A]0 and [B]0 respectively. (a) Give an argument to show that [B] = [B]0 + [A] − [A]0 . (b) Solve the differential equation d[A] = −k2 [A][B] dt to obtain the result [A] =

c[A]0 exp (−ck2 t) c + [A]0 − [A]0 exp(−ck2 t)

where c = [B]0 − [A]0 .

Solution (a) In the reaction the same number of molecules of [A] and [B] combine to give the product so the change in the concentration of [A] is the same as the change in the concentration of [B], that is [B] − [B]0 = [A] − [A]0 (b) Using [B] = c + [A] with c = [B]0 − [A]0 we have d[A] = −k2 [A](c + [A]) dt Solving we have Z d[A] = −k2 dt [A](c + [A]) Z 1 1 1 d[A] = −k2 t + C − c + [A] [A] c [A] = −k2 ct + C ′ ln c + [A] [A] = C ′′ e−k2 ct c + [A] cC ′′ e−k2 ct [A] = 1 − C ′′ e−k2 ct Z

where C, C ′ and C ′′ are different constants. Now when t = 0 [A] = [A]0 so [A]0 =

2

cC ′′ 1 − C ′′

but it is actually easier to go back to C ′ and notice ln

[A]0 [A]0 = ln = C′ c + [A]0 [B]0

and hence C ′′ = [A]0 /[B]0 . Substituting in the generals solution and multiplying numerator and denomonator by [B]0 we obtain the desired result.

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