Ngô Quốc Anh

February 16, 2010

ODE: The method of undetermined coefficients

Filed under: PDEs — Ngô Quốc Anh @ 12:59

Happy Lunar New  Year!

This method is more limited in scope; it applies only to the special case of

\displaystyle y'+p(t)y=g(t)

where p(t) is a constant and g(t) has some special form. The advantage of the method is that it does not require any integrations and is therefore quick to use. The homogeneous equation

\displaystyle y'+\lambda y=0

has the solution

\displaystyle y_h=Ce^{-\lambda t}.

To solve the inhomogeneous equation

\displaystyle y'+\lambda y=g(t),

it suffices to find one particular solution y_p(t). If y_p(t) is any particular solution, then the general solution is

\displaystyle y(t)=y_p(t)+Ce^{-\lambda t}.

The idea behind the method of undetermined coefficients is to look for y_p(t) which is of a form like that of g(t). This is possible only for special functions g(t), but these special cases arise quite frequently in applications.

Case 1. We start with the case where g(t) is an exponential

\displaystyle g(t)=A e^{\alpha t}.

We look for y(t) in a similar form

\displaystyle y(t)=a e^{\alpha t}.

This leads to

\displaystyle y'=a\alpha e^{\alpha t}, \quad y'+\lambda y=(\alpha+\lambda)ae^{\lambda t}.

So the differential equation becomes

\displaystyle (\alpha+\lambda)ae^{\lambda t} = Ae^{\alpha t}.

We can solve this to find

\displaystyle a=\frac{A}{\alpha+\lambda}.

This leads to the particular solution

\displaystyle {y_p}(t) = \frac{A}{{\alpha + \lambda }}{e^{\alpha t}},

and the general solution

\displaystyle y(t) = \frac{A}{{\alpha +\lambda }}{e^{\alpha t}} + C{e^{ - \alpha t}}.

Case 2. If y is a polynomial of degree n, then y' is a polynomial of degree n-1. If g is a polynomial, we can therefore look for polynomial solutions. Consider

\displaystyle y'+2y=t^2.

The right hand side is a polynomial of degree 2, so we look for a solution in the same form y=at^2+bt+c. This leads to y'=2at+b, and

\displaystyle y'+2y=2at^2+(2a+2b)t+b+2c=t^2.

To satisfy this, we want to set

\displaystyle 2a=1, \quad 2a+2b=0, \quad b+2c=0.

This leads to a=\frac{1}{2}, b=-\frac{1}{2}, c=\frac{1}{4}. So a particular solution is

\displaystyle y_p=\frac{t^2}{2}-\frac{t}{2}+\frac{1}{4}.

The general solution is

\displaystyle y=\frac{t^2}{2}-\frac{t}{2}+\frac{1}{4}+Ce^{-2t}.

We note that the solution

\displaystyle y(t) = \frac{A}{{\alpha +\lambda }}{e^{\alpha t}} + C{e^{ - \alpha t}}

breaks down if \alpha=-\lambda, since it would involve a division by zero. More generally, if the equation reads

\displaystyle y'+\lambda y=g(t),

and g(t)=e^{\alpha t}P_n(t), with P_n(t) an nth degree polynomial, then we can find a particular solution

\displaystyle y_p(t)=e^{\alpha t} Q_n(t),

where Q_n(t) is some other nth degree polynomial as long as \alpha \ne -\lambda. If, on the other hand, \alpha=-\lambda, we have to modify the procedure. The modification is simply to include an extra factor t in the solution. That is, instead of setting

\displaystyle y_p(t)=e^{\alpha t} Q_n(t),

you set

\displaystyle y_p(t)=t e^{\alpha t} Q_n(t).

Source: http://www.math.vt.edu/people/renardym/class_home/firstorder/node2.html

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