ENGI 9420 Engineering Analysis
Mid Term Test Solutions

Two methods for a solution are shown here.

Separation of Variables:

Pursuing the non-singular solution,

Imposing the initial condition,

But the singular solution also satisfies the initial condition!
Therefore the complete solution is

Conversion to Exact Form (using an integrating factor):

Try to find an integrating factor as a function of   x   only:

Try to find an integrating factor as a function of   y   only:



Multiplying the ODE by   I(y)   will render the ODE exact, unless , when a division by zero occurs.
is a singular solution to the ODE.
The exact ODE is     2x dx – dy = 0

Imposing the initial condition,

But the singular solution also satisfies the initial condition!
Therefore the complete solution is

By any valid method, find the function   y(t)   whose Laplace transform is

Hence (or otherwise) solve the initial value problem

where   H(t – a)   is the Heaviside (or unit step) function  

Two methods for finding     are presented here.

Partial Fractions:

Matching coefficients:

Integration Property:

The factor   s²   in the denominator of   G(s)   corresponds to two successive integrations in t-space:

OR one may quote the inverse Laplace transform directly from the tables

Using the second shift theorem,

One form of the second shift theorem is

Taking the Laplace transform of the initial value problem,


which is precisely the inverse Laplace transform problem solved above.   Therefore

Find the MacLaurin series solution of the ordinary differential equation

with arbitrary initial conditions   y(0) = a,   y' (0) = b , as far as the term in x⁷.

Substitute the MacLaurin series solution     into the ODE:

Shift the indices on the first summation:

Combining the three summations together,

Matching coefficients of successive powers of   x :



For   n > 2   we have the recurrence relation



Also note that   y(0) = a₀ = a   and   y'(0) = a₁ = b
Therefore the series solution to the ODE, as far as the term in x⁷, is

This series can also be written as

OR

The Maclaurin series for   y(x)   is also

The two initial conditions provide the first two terms of the series immediately:
y(0) = a   and   y'(0) = b
Rearrange the ODE:

Differentiate the ODE repeatedly:



At this point a recurrence relationship develops:

But   y'''(0) = 0   so that all odd-order derivatives of   y(x)   (except the first) are zero at   x = 0.
Also   y^IV(0) = 1 – a,   so that all higher even order derivatives at   x = 0   also include the factor (1 – a).
These relationships are needed for the bonus question.
The remaining term needed for this question is

The series then follows:

BONUS QUESTION
Find the condition(s) on   a   and/or   b   under which the series solution has only a finite number of non-zero terms and write down that finite series solution.

Note first that   .   Recall the recurrence relations
  or  
Therefore the coefficients of all the odd powers of x (except 1) are zero and
the coefficients of every even power of x from 4 onwards share the same factor of (1 – a).

Therefore the series has only a finite number of non-zero terms if and only if

when the series is

(where b remains arbitrary).

Additional Note:   The complete series can be re-written as

The cubic equation
             
is a special case (e = 0.01) of the perturbation
             
of the simple cubic equation
             
Find the perturbation series, as far as the term in e², for the root near x = –1 and hence find the solution of nearest to x = –1, correct to six decimal places.

Solution by Expansion:

Let the perturbation series be

Substitute this series into the cubic equation:



Rearrange the coefficients of   e ⁰, e ¹, e ²   into a tabular display:

Equating coefficients of e ⁰:
, (which is the only real solution to the unperturbed equation).
Equating coefficients of e ¹:

Equating coefficients of e ²:

Therefore the perturbation series is

The solution to the equation     is therefore

Therefore the solution to the perturbed cubic equation, correct to six decimal places, is

x = –1.010 067

OR

Solution by Iteration:

An obvious iteration scheme is

Use the standard binomial expansion

(valid for   | z | < 1)
and start the iterations with the solution to the unperturbed cubic equation  
For   x₁   discard all terms in the binomial expansion beyond   e ¹ :

For   x₂   discard all terms in the binomial expansion beyond   e ² :

This result must be verified by finding   x₃ :
For   x₃   discard all terms in the binomial expansion beyond   e ³ :


This confirms the series as far as the term in   e ².
The same answer of   –1.010 067   then follows.

[This answer is indeed correct to six decimal places.
The exact answer is   –1.010 066 775 168...]