Method of undetermined coefficients Problems

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The discussion focuses on solving differential equations using the method of undetermined coefficients and variation of parameters. For the equation 3y''-6y'+30y=15sinx+exp(x)tan(3x), participants suggest separating the equation into two parts, applying variation of parameters to the exponential term and undetermined coefficients to the sine term. The user shares their work on finding the homogeneous solution and specific solutions, detailing calculations for coefficients A and B. They also mention the complexity of the equations and recommend using computational tools like Mathematica for assistance. The conversation emphasizes the importance of understanding the method applied to each part of the equation for effective problem-solving.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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