posted by Sara on .
Find the complete solution to the 2nd ODE:
Start with solving for the complementary solution to the homogeneous equation
the characteristic equation is
which is an algebraic equation which can be factorized into
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:
For this, we will need to assume that the particular solution yp is as follows:
Now to find the constants C,D,E,F,G, we need to calculate
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:
which when solved, gives
Now tackle C, D and G:
Compare coefficients of t² yields
Substitute C and compare coefficients of t gives D=-6
and consequently coefficient G=14.
The general solution is therefore:
=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: