Posted by Sara on .
Find the complete solution to the 2nd ODE:
2y''+3y'+y=t^(2)+3sin(t)

CalculusODE 
MathMate,
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 righthand 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:
3FE=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} +
t²6t 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. 
Calculus 
MathMate,
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