# Homework Help: Method of undetermined coefficients Problems

1. Nov 16, 2009

### drew1435

been working on these all week some help please..

9) 3y''-6y'+30y= 15sinx+exp(x)tan(3x)

10) y''-2y'+y= 4x^2 -3+ Exp(x)/x

2. Relevant equations

Variations of parame
Method of undetermined coefficients

thoughts:

i think that you need to split the eqtuaion into 2 parts. Using variation of parameters on 3y''-6y'+30y=exp(x)tan(3x).. and using method of undetermined coefficients on 3y''-6y'+30y=15sinx.. i have tried this approach on both problems with no luck.

Last edited: Nov 16, 2009
2. Nov 16, 2009

### HallsofIvy

It's impossible to tell what you should do without knowing what you did and seeing where you ran into trouble. What did you do?

3. Nov 17, 2009

### drew1435

sorry here is the work i did for #9

Generic solution:
y = e^Ax
y' = Ae^Ax
y'' = A^2 e^Ax

3A^2 - 6A + 30 = 0
3(A^2 - 2A + 10) = 0
A = [2 +/- sqrt(4 - 40) ]/2 = 1 +/- sqrt(-9) = 1 +/- 3i

y = C1 e^(1 +/- 3i)x + C2 e^(1 +/- 3i)x
y = C1 e^(1 +/- 3i)x + C2 e^(1 +/- 3i)x
y = C1 e^x (cos 3x + i sin 3x) + C2 e^x (cos 3x - i sin 3x)
y = C1' e^x cos 3x + C2' i e^x sin 3x

specific soluition using undetermined coefficients.

15sinx
If we plug in y = A sin x, we get:

3y''-6y'+30y= 27 sin x - 6 cos x

And if we plug in y = B cos x, we get:

3y''-6y'+30y= 27 cos x + 6 sin x

With a little vector addition:

27A + 6B = 15; 9A + 2B = 5
-6A + 27B = 0; A = 9/2 B

81/2 B + 2B = 5
85/2 B = 5
B = 10/85 = 2/17
A = 9/17

So y = 9/17 sin x + 2/17 cos x is the specific solution that will generate 15 sin x.

I can't quite get the third piece

4. Nov 18, 2009

### jakncoke

This is a damn hairy diffeq. For 9, use variation of parameters. First calculate the homogeneous equation. 3y''-6y'+30y=0. since this has constant coeff, assume that the answer is of the form y=e^(mx) and plugging it in u get a auxiliary equation of the form

3m^2-6m+30=0. Solving it u get the homogenous eqn y= c[1]*Exp[x]Sin[3x]+c[2]*Exp[x]Cos[3x].

now that we have the homogeneous eqn, we can use variation of parameters to calculate v1.
where v1 = Integral[(-y2R(x))/W] for the Wronskian we get 3Exp[2x]. so

v1 = -Integral[{(Exp[x]Cos[3x])*(15Sin[x]+Exp[x]Tan[3x])}/3Exp[3x]]
solving that puppy out to take u a while but after Integrating you should get

v1 = 1/816 E^(4 x) (204 Cos[2 x] - 24 E^x Cos[3 x] - 255 Cos[4 x] -
408 Sin[2 x] + 40 E^x Sin[3 x] + 255 Sin[4 x])

then multiply v1*y1 to get the first part of the particular soln.

then find v2, multiply by y2 etc...

b/w i used mathematica to calculate v1 and i suggest u do the same for the others unless ur teacher specified otherwise. gl.

also same process as above for 10.

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