ENGR 2422 Engineering Mathematics 2

Brief Notes on Chapter 4


Second Order Linear Ordinary Differential Equations

Contents:

4.1Complementary Function

4.2Particular Solution (Undetermined Coefficients)

4.3Particular Solution (Variation of Parameters)

4.4Higher Order Linear Ordinary Differential Equations


4.1��� Complementary Function

 

Of the second (and higher) order ordinary differential equations, only linear equations with constant coefficients will be considered in this chapter:

The homogeneous equation associated with this ODE is

The principle of superposition of solutions of the homogeneous equation is valid because it is linear.�� That is, ify = u(x)and�� y = v(x)are both solutions of the homogeneous ODE, then so also is��

y = c1 u(x)+c2 v(x) , wherec1andc2are any constants.

Adding any solution of the homogeneous ODE to a particular solution of the original ODE generates another solution of the original ODE.

 

Thus the general solution of

can be partitioned into two parts:

the complementary function (which is the general solution of the associated homogeneous ODE) and a particular solution.

 

If�� y = elxis a solution to the homogeneous ODE, then the auxiliary equation follows:

l2+P l+Q=0

Three cases result, (in whichc1, c2, c3, c4 are arbitrary constants of integration):

If the rootsl1, l2of the auxiliary equation are real and distinct, then the complementary function is

If the rootsl1, l2of the auxiliary equation are real and equal, then the complementary function is

If the rootsl1, l2of the auxiliary equation are the complex conjugate pair a bj, then the complementary function is

This last case can also be written in the forms

yC=A eax cos (dx - d)���� or���� yC=A eax sin (dx - d)

where�� Aanddare arbitrary constants of integration.


An example of a linear homogeneous second order ordinary differential equation with constant coefficients is the model of the extensionsof a spring of massmmoving in one dimension under the influences of a restoring force-csand a drag force-bv (wherevis the speed of the free end of the spring):

The auxiliary equation is

The behaviour of the spring then depends on the relative values ofb , candm.

 

the system is under-damped and exhibits damped oscillations:

s(t)=A e-at sin (kt - d)

where

 

the system is over-damped and exhibits no oscillations at all, just exponential decay:

where

 

the system is critically damped and passes through zero at most once before settling asymptotically to zero as rapidly as possible:

s(t)=(A + Bt) e-bt/(2m) .

 

 

Ifb = 0(no friction at all), then the system is totally undamped and exhibits simple harmonic motion:

s(t)=A sin (kt - d)


The Operator Method

 

The homogeneous ordinary differential equation

can also be written, using differential operators, in the form

where-k1and-k2�� are the solutions to the auxiliary equationl2+P l +Q=0.

This gives rise to the linked pair of first order linear ordinary differential equations

and.

It is shown in class that the general solution to this pair of ODEs is the complementary function introduced on page 27.


4.2��� Particular Solution (Undetermined Coefficients)

 

The general solution to

is the sum of the complementary function and any one solution (the particular solution) that we can find to the original inhomogeneous ODE.

 

If the functionR(x)does not contain any part of the complementary function, then assume that the particular solutionyP(x)is of the same form asR(x).

 

IfR(x) = ekx ,

then try�� yP = c ekx , with cto be determined.

 

IfR(x) = (a polynomial of degree n),

then try yP = (a polynomial of degree n), with all (n + 1) coefficients to be determined.

 

IfR(x) = (a multiple ofcos kxand/orsin kx),

then try yP = c cos kx + d sin kx

 

This method can be extended to cases where R(x) =(a sum and/or product of the functions above).

 

But:if part (or all) of yP is included in the C.F., then multiply yP byx.


Example 1��� Find the general solution to the ODE

y"+2 y'-3 y=x2+e2x

 

[The details will be covered in class.]

The auxiliary equation is�� l2+2 l-3=0

which has the solutions�� l=-3 or +1.

The complementary function is therefore�� yC(x)=A e-3x+B ex

 

R(x) =x2+e2xwhich contains neither e-3x norex .

Therefore try a particular solution of

yP(x)=ax2+bx+c+de2x

Upon settingyP"+2 yP'-3 yP=x2+e2x

and matching coefficients ofx2 , x1 , x0and e2x , we find the values ofa, b, c and d.

 

The general solution isy(x)=yC(x)+yP(x):


Example 2

 

A model of the simple series LRC circuit leads to the differential equation

where the constantsR, L, Care the resistance, inductance and capacitance respectively,E(t)is the applied electromotive force,t is the time andI(t)is the resulting current.

The details will be covered in class.��

With the constantDdefined to be the negative of the discriminant, ,

the complementary function generates a transient term (that decays to zero as t )

The form of the particular solution depends on the applied electromotive forceE(t).

IfE(t)is constant, then the particular solution isIP(t) = 0.

IfE(t)is sinusoidal, then the particular solution provides a steady state component to the current that is also sinusoidal.�� In particular, ifE(t)=Eo sin wt, then


Example 3

Find the complete solution of the ODE

y"+2 y'+y=e-x��� ,��� y(0) = y' (0) = 1

The auxiliary equationl2+2 l+1=0has the repeated rootsl = -1, -1.

The complementary function is�� yC=(Ax + B) e-x .

Bothy =e-xand y =x e-xare included in the complementary function.

Therefore tryyP =a x2 e-x:

y"P+2 y'P+yP�� =e-x ���

((2a - 4ax + a x2) + (4ax - 2a x2) + (a x2) e-x=1 e-x

�� ((a - 2a + a) x2 + (- 4a + 4a) x + (2a) e-x=1 e-x

�� a=1/2

Therefore the general solution is

Now impose the initial conditions on this general solution:

y(0)=(0+0+B) e0=1��� �� B=1

y' (x)=(x + A -x2 - Ax - B) e-x

��� y' (0)=(0 + A - 0 - 0 - 1) e0=1��� �� A - B= 1��� �� A = 2

Therefore the complete solution is

 

 

 

Note that a complete solution requires additional information (often in the form of initial conditions).�� Two pieces of information are needed in order to evaluate both arbitrary constants of integration.�� However, do not substitute these conditions into the complementary function; wait until the general solution has been obtained.

 

 


4.3��� Particular Solution (Variation of Parameters)

 

The method of variation of parameters is a more general method for finding the particular solution.�� It is successful even in some cases where the method of undetermined coefficients fails.�� However, where both methods are available, the method of undetermined coefficients is generally faster to use.

 

In class the following will be derived:

If the complementary function for the ODE

is�� yC(x)=C1y1(x)+C2y2(x) , then the particular solution is

yP(x)=u(x) y1(x)+�� v(x) y2(x) , where

 

The functionW(x)is known as the �Wronskian� ofy1 andy2.

 

Three examples will be covered in class.�� Example 2 is a case that cannot be solved by the method of undetermined coefficients:

y" + y=tan x

R(x) = tan xis not one of the standard forms.

Upon evaluation, one can see that the particular solution is not a simple linear combination of any trigonometric functions:

yP(x)=-(cos x) ln | sec x + tan x |

 


Modified Method of Undetermined Coefficients

 

If part of the complementary function,y1, is included in the functionR(x), then try

yP=f (x) y1as a particular solution.�� Substitute into the ODE and solve for f (x).

An example will be demonstrated in class.


4.4��� Higher Order Linear Ordinary Differential Equations

 

The nth order ordinary differential equation

can be solved as follows.

Form the auxiliary equation

ln+a1ln-1 + ... +an-2l2 +�� an-1l1 + an=0

Find all n values forl.

Form the complementary functionyC, which will be a linear combination of

(except for repeated roots).��

Complex conjugate pairs can be re-written in terms of sine and cosine functions.

 

Find a particular solution yP (by inspection, undetermined coefficients, or variation of parameters, as extended to this higher order equation).

 

Write down the general solution�� y=yC+yP.

ninitial and/or boundary conditions will be needed at this stage to evaluate all of the n arbitrary constants of integration.


Example[Demonstrated in more detail in class.] �� Find the general solution of

Auxiliary equation:

l5 + 2l4 - 3l3 - 4l2 + 4l=0

�� ...�� �� l(l-1)2(l+2)2=0

�� l=0, 1, 1, -2, -2.

Complementary function:

yC=A+(Bx + C) ex +(Dx + E) e-2x

Particular solution:

Cannot try�� yP=ax + bbecause a constant is included in the complementary function.

Therefore try��� yP=(ax + b) x=ax2 + bx

Upon substituting this into the ODE, we find thata = 1andb = 2.

Therefore the general solution is

y=A+(Bx + C) ex+(Dx + E) e-2x+ x2+ 2x

 

Five initial conditions would be sufficient to evaluate the arbitrary constantsA, B, C, D and E.


[Cauchy-Euler ODEs will not be covered in 2001 Winter.]


END OF CHAPTER 4


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Created 2001 02 28 and modified 2001 02 28 by Dr. G.H. George