Method of undetermined coefficients Problems

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Homework Help Overview

The discussion revolves around solving differential equations using the method of undetermined coefficients and variation of parameters. The specific equations being analyzed include a non-homogeneous linear differential equation with sinusoidal and exponential terms.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore splitting the equations into parts for different solution methods. There is an attempt to derive the homogeneous solution and apply the method of undetermined coefficients to specific terms. Questions arise regarding the steps taken and difficulties encountered in the process.

Discussion Status

Some participants have provided detailed calculations and suggested using variation of parameters for one of the equations. Others have requested clarification on the original poster's work to better understand the challenges faced. There is no explicit consensus on the best approach yet, but various methods are being discussed.

Contextual Notes

Participants note the complexity of the differential equations and the need for specific techniques, such as the use of software tools for calculations. There is an emphasis on understanding the homogeneous solutions before proceeding with particular solutions.

drew1435
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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)/x2. Homework 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.
 
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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?
 
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
 
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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