Elementary Differential Equations: the method of undetermined coefficients

October 15, 2010

Elementary Differential Equations: the method of undetermined coefficients

Filed under: applied mathematics, differential equations — collegemathteaching @ 1:45 am

Just a brief review: if we have a second order linear differential equation with constant coefficients $y^{''} + by' + ay = f(t)$ we obtain the solution in the following manner: first we solve the related homogeneous differential equation $y^{''} + by' + ay = 0$ to obtain $y_{h}$ Then, if $f$ is the “right” kind of function, one can make a guess at what a particular solution $y_{p}$ would be up to a constant multiple. One then inserts their candidate for the particular solution into the differential equation to solve for the constant.

What is the “right” kind of function? Basically these are the classes of functions that form a closed class under differentiation. Examples of such closed classes would be exponential functions, sines and cosines (together), polynomials, and possibly multiples of these.

Here is an example to refresh your memory: suppose $y^{''}+y =e^{t}$ Then $y_{h} = Asin(t)+Bcos(t)$ and we assume $y_{p} = De^{t}$
Then putting back into the differential equation we obtain $2De^{t}=e^{t}$ which implies $y_{p}=(1/2)e^{t}$ So it follows that the general solution is $y = Asin(t)+Bcos(t) +(1/2)e^{t}$

Trouble arises when the forcing function is itself part of the homogeneous solution; example $y^{''}+y = sin(t)$
We usually tell our students to multiply a candidate for the particular solution by $t$ and try again; e. g. in this example we’d attempt $y_{p} = t((r)cos(t)+(s)sin(t))$ which works out to $y_{p}=(1/2)tcos(t)$ . But when does this work?

Let’s see: Suppose we have $y^{''} + by' + ay = y_{h1}$ and we attempt $y_{p} = Aty_{1} + Bty_{2}$ where $y_{1}$ and $y_{2}$ are two linearly independent homogeneous solutions. Put these into the differential equation and do some algebra and we obtain:
$2Ay'_{1} +2By'_{2} +aAy_{1}+aBy_{2} = y_{h1}$

Case One: $y_{h}=\alpha e^{rt}\cos (wt)$ $+$ $\beta e^{rt}\sin (wt)$
(note: we can let $r=0$ if needed)

So $y_{1}=Ate^{rt}\cos (wt),y_{2}=Bte^{rt}\sin (wt)$ and then $2y_{h}^{\prime }+ay_{h}$ becomes:

$((2r+a)B-2wA)e^{rt}\sin (wt)+(2wB+(2r+a)A)e^{rt}\cos (wt).$ Equating
coefficients with the forcing function can lead to a matrix system whose
left hand side is:

$\left[ \begin{array}{cc}-2w & 2r+a \\ 2r+a & 2w\end{array}\right] \left[ \begin{array}{c}A \\ B\end{array}\right]$ The coefficient matrix has rank 2 even if $r=0$ and $a=0.$ Hence
the system can always be solved.

Case Two $y_{h}=ae^{r_{1}t}+be^{r_{2}t},r_{1} \neq r_{2}$
Then $y_{1}=Ate^{r_{1}t},y_{2}=Bte^{r_{2}t}$ and then $2y'_{h}+ay_{h}$ becomes:
$A(2 r_{1}+a)e^{r_{1}t} + B(2 r_{2} + a)e^{r_{2}t}$ which is always solvable unless $2r_{1}+a=0$ or $2r_{2}+a=0.$ In either case, this implies that $y_{1}$ or $y_{2}$ are themselves part of the homogeneous solution. This puts us in

Case Three: $y_{h}=ae^{rt}+bte^{rt}$
Here: $ty_{h}$ will NOT work as the first term is part of the homogeneous solution. This is the only case where $t^2y_{h}$ is required.
So attempt $y_{p}=t^2y_{h}=t^2(A+tB)e^{rt}$
Putting into the differential equation we end up with $2(at+1)y_{h}+4ty'_{h}$
which upon substituting $y_{h}=t^2(A+tB)e^{rt}$ , becomes:
$e^{rt}(B(2a+4r)t^2+(2aA+4Ar+6B)t+2A)$ .
But remember that $2r=-a$ as we were in the “real repeated root situation” so this becomes:
$e^{rt}(6Bt+2A)$
Which, of course, can always be solved. Note that $r$ plays no role in the coefficients.

Example: if $y^{''}+2ry^{'} + r^{2}y = e^{-rt}+te^{-rt}$ Then one can easily check that a particular solution is given by $y_{p} =(1/2)t^{2}e^{-rt}+(1/6)t^3e^{-rt}$

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[...] didn’t post much today but I did write this post about the “method of undetermined coefficients” for solving inhomogeneous second order [...]

Pingback by 14 October 2010 pm « blueollie — October 15, 2010 @ 1:49 am

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College Math Teaching

October 15, 2010