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EBL Maths Questions | Week 19 David Silvester | 7 March 2012

1. Solve analytically the following differential equations:

(a)dy

dx= x2y ; y = 1 when x = 0.

(b) dydx

= y2 sinx ; y = 1 when x = /2.

(c)dy

dx= x(4 + y2) ; y = 2 when x = 0.

Solution

We need to Separate and Integrate in each case:

(a) Separating gives

dyy

=

dxx2

, integrating gives loge y = 13x

3 + C. Usingthe initial condition y(0) = 1 we see that C= 0.

(b) Separating gives

dyy2

=

sinx dx, integrating gives 1

y= cos x + C. Using

the initial condition y(/2) = 1 we see that C= 1. Hence y =1

(1cos x) .

(c) Separating gives

dy4+y2

= x dx, integrating gives 12 tan

1(y2) = 12x

2 +C.

Using the initial condition y(0) = 2 we see that C= /8.Hence tan1(y2 ) = x

2 + 4 and y = 2 tan(4 x

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]

dt= k2[A][B]

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) The change in concentration ofB must be equal to the change in concentrationofA. This means that [B] [B]0 = [A] [A]0.

(b) To simplify the notation we let y = [A], c = [B]0 [A]0 so that y + c = [B].The problem is now to find y(t) satisfying

dy

dt= k2 y(y + c).

Separating gives

dyy(y+c) =

k2 dt, taking partial fractions gives

1y(y+c) =

1cy 1

c(y+c) and then integrating gives1c

loge y 1c

loge(y + c) = k2t + C.

Rearranging this gives

logey + c

y= ck2t + K.

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We then set t = 0 so that y = [A]0, and y + c = [B]0 to get K = loge[B]0[A]0

.Substituting this above gives

loge

[A]0[B]0

1 +

c

y

= ck2t

[A]0[B]0

1 +

c

y

= exp(ck2t).

Rearranging this final expression gives the required result:

y = [A] =c[A]0 exp(ck2 t)

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

3. Solve the following linear ODEs by finding an integrating factor:

(a)dy

dx+ y = exp(x) ; y = 2 when x = 0.

(b)dy

dx+ y cosx = cos x ; y = 1 when x = 0.

(c)dy

dx +y

x = sinx ; y = 1 when x = 0.

Solution

(a) The integrating factor is exp(p dx). In this case we have that p = 1 so

the I.F is simply ex. Multiplying both sides of the ODE by the I.F. we obtain

exdy

dx+ exy = 1

d

dx(yex) = 1.

Integrating both sides then gives yex =

1 dx = x + C. Using the initialcondition y(0) = 2 we see that C= 2. Hence y = (2 + x)ex.

(b) In this case we have that p = cos x. Since

cos x dx = sinx we find thatthe I.F. is esinx. Multiplying both sides of the ODE by the I.F. we obtain

esinxdy

dx+ yesinx cosx = esinx cos x

d

dx(yesinx) = esinx cosx.

Integrating both sides then gives yesinx =esinx cosx dx = esin x + C. Using

the initial condition y(0) = 1 we see that C= 0. This means that the solutionis simply the constant function y = 1.

(c) In this case we have that p = 1/x. Since

dxx

= loge x we see that the I.F.is simply x. Multiplying both sides of the ODE by the I.F. we obtain

xdy

dx+ y = x sinx

d

dx(yx) = x sinx.

Integrating both sides then gives yx =x sinx dx = sinx x cos x + C.

Up to here everything is fine. However looking at the initial condition y(0) = 1we run into a BIG problem; namely, setting x = 0 we find that C= 0 and sothe solution y = sinx

x cos x does not depend on the given value of y(0)! In

fact the solution turns out to be inconsistent with the initial condition sincethe limit of sinx

xwhen x approaches zero is the value 1. This means that

y(0) = 0 = 1. The moral is that not every ODE problems has a solution.

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