solve y"-3y'+2y=cosx

Question

Solve the differential equation

Answer 1

y"-3y'+2y=cos(x).

The characteristic solution is found by solving y"-3y'+2y=0.

Replace this with the quadratic r²-3r+2=(r-2)(r-1)⇒y_c=Ae^2x+Be^x, where A and B are arbitrary constants.

Now we need a particular solution of the form y_p=asin(x)+bcos(x), where a and b are specific constants to be found. We assume this form for the solution because we have the cosine function cos(x).

y_p'=acos(x)-bsin(x); y_p"=-asin(x)-bcos(x). Substitute in the DE:

-asin(x)-bcos(x)-3(acos(x)-bsin(x))+2(asin(x)+bcos(x))=

sin(x)(-a+3b+2a)+cos(x)(-b-3a+2b)≡cos(x) by matching coefficients:

a+3b=0 and -3a+b=1. So a=-3b⇒10b=1, b=0.1⇒a=-0.3.

So y_p=-0.3sin(x)+0.1cos(x).

[y_p"-3y_p'+2y_p=(0.3sin(x)-0.1cos(x))-3(-0.3cos(x)-0.1sin(x))+2(-0.3sin(x)+0.1cos(x))=

sin(x)(0.3+0.3-0.6)+cos(x)(-0.1+0.9+0.2)=cos(x), so y_p is correct.

y=y_c+y_p=Ae^2x+Be^x-0.3sin(x)+0.1cos(x) is the complete solution.