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April 16, 2014

April 16, 2014

Posted by **Sara** on Thursday, July 18, 2013 at 6:15pm.

2y''+3y'+y=t^(2)+3sin(t)

- Calculus-ODE -
**MathMate**, Friday, July 19, 2013 at 8:10amStart 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.

- Calculus -
**MathMate**, Friday, July 19, 2013 at 8:11amIf 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

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