Posted by Sara on Thursday, July 18, 2013 at 6:15pm.
Start with solving for the complementary solution to the homogeneous equation
2y"+3y'+y=0,
the characteristic equation is
2r²+3r+1=0
which is an algebraic equation which can be factorized into
(2r+1)(r+1)=0
or
r={-(1/2), -1}
The complementary solution is therefore:
yc=Ae^{-(x/2)} + Be^{-x}
The next step is to find the particular solution for the right-hand side:
y²+3sin(t)
For this, we will need to assume that the particular solution yp is as follows:
yp=Ct²+Dt+Esin(t)+Fcos(t)+G
Now to find the constants C,D,E,F,G, we need to calculate
2yp"+3yp'+yp=t²+sin(t)
and equate coefficients.
We end up with:
G+(-3sin(t)-cos(t))F+(3cos(t)-sin(t))E+(t+3)D+(t^2+6*t+4)C = t²+sin(t)
which can solved for C,D,E,F,G by comparing coefficients and successive substitution:
Sum coefficients of sin(t)=1,
sum of coefficients of cos(t)=0:
-3F-E=1
-F+3E=0
which when solved, gives
E=-1/10, F=-3/10.
Now tackle C, D and G:
Compare coefficients of t² yields
C=1
Substitute C and compare coefficients of t gives D=-6
and consequently coefficient G=14.
The general solution is therefore:
y(t)=yc+yp
=Ae^{-(t/2)} + Be^{-t} +
t²-6t -sin(t)/10 -3cos(t)/10+14
Finally, we check that
2y"(t)+3y'(t)+y(t) equals t²+sin(t)
which confirms that the solution is correct.
If you need help with finding the standard forms of the particular solution, check your course notes, your textbook, or the following link:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx