Question

allreal solutions of the DE y"+2ay'+by=cosx are periodic if

Answer 1

A particular solution of the DE would be

yp=cos(x)−bcos(x)−2asin(x)(1+2a2+2aa2−b−−−−−√−b)(−1−2a2+2aa2−b−−−−−√+b)yp=cos⁡(x)−bcos⁡(x)−2asin⁡(x)(1+2a2+2aa2−b−b)(−1−2a2+2aa2−b+b).

And the complimentary solution is

yc=c1e(−a−a2−b√)x+c2e(−a+a2−b√)xyc=c1e(−a−a2−b)x+c2e(−a+a2−b)x.

So the solution of the DE is

y=c1e(−a−a2−b√)x+c2e(−a+a2−b√)x+cos(x)−bcos(x)−2asin(x)(1+2a2+2aa2−b−−−−−√−b)(−1−2a2+2aa2−b−−−−−√+b)y=c1e(−a−a2−b)x+c2e(−a+a2−b)x+cos⁡(x)−bcos⁡(x)−2asin⁡(x)(1+2a2+2aa2−b−b)(−1−2a2+2aa2−b+b).

Now as you can see if a2≥ba2≥b, then all solutions are real and has a periodic part and a non-periodic part. To cancel out the non-periodic part there are two ways. If the arbitrary constants c1c1 and c2c2 becomes zero then the solution would be periodic. But that would not be a general solution. Also if the exponents becomes zero then again the solution would be periodic as the exponential part would be reduced to a constant. This would be a general solution. So for the exponents to be zero the following is required,

−a−a2−b−−−−−√=0−a−a2−b=0 and −a+a2−b−−−−−√=0−a+a2−b=0.

It is easy to notice that both of them can not be zero at the same time unless both aa and bbare zero. When both aa and bb are zero the DE reduces to

y′′=cos(x)y″=cos⁡(x)

which is very easy to solve and the general solution is

y=−cos(x)y=−cos⁡(x).

I used the method of undetermined constants. Feel free to correct me if any computational error has occurred in my calculations.