How to solve differential equation: y''''+3y'''+2y''=5x-e^(-x)*cos2x where y=y(x) [first one is 4th differential, second one is 3rd differential and third one is second differential] p.s. Any great materials on solving higher order differential equations? Thanks
\[y''''[x] + 3 y'''[x] + 2 y''[x] = 5 x - Cos[2 x] * e^{-x}\]
I would suggest that you use a substitution \(u=y''\). Then solve the associated homogeneous equation, which should be very easy.
The hardest part is probably finding the particular solution. There are number of ways to finding it. I would either use the annihilator approach or superposition approach.
Do you want to see this out in gory detail?
that would be kinda likable, but I won't cry, if I have to deal with it by myself.
Well, do you know how to solve the homogenous part?
In other words the: \[y^{(4)}(x)+3y'''(x)+2y''(x)=0\]?
Be back in 20 btw then I can finish helping you, dinner haha
Hmm... Im thinking of substituting y''=u (as stated above) and then the solution should be proportional to e^(C*x) where C is some constant. Then solve equation for u. substitute back for u=y''
if I'm right \[y''=C_11*e^{-2x}+C_2*e^{-x}\]
The substitution I suggested above will simplify the problem, esp when dealing with the particular solution. You can solve the problem without it though.
That could work, for convenience. I can show you either method. Which would you prefer? If you make that substitution u=y'', would you convert it to a system? Or would you just solve for u then solve another second order diffeq? I would NOT do the substitution (no offense Mr. Math) but its completely up to you.
couldn't I just integrate two times when I get y''=... wait.. then I will have the solution in form y^2=...
and I don't know, whether second order equation is good, considering that this solution is only the homogeneous part
any help? malevolence19?
Okay. To do the homogenous part: \[y^{(4)}(x)+3y'''(x)+2y''(x)=0; \rho^4+3 \rho^3+2 \rho^2=0 \implies \rho^2 (\rho^2+3\rho+2)=0\] Factoring this we get: \[\rho^2 (\rho +2)(\rho +1)=0 \implies \rho=0,0, -2,-1\] Which means our homogenous solution is: \[y_H(x)=c_1 e^{0x}+c_2 x e^{0x}+c_3 e^{-2x}+c_4 e^{-x}=c_1+c_2 x+c_3 e^{-2x}+c_4 e^{-x}\] Do you follow me up to here?
Yes, seems legit.
Okay. How do you know how to find your particular solution? I plan on using the method of undetermined coefficients as I find it easiest.
undetermined coefficients being... (sorry, english is not my native language.)
You select a family of solutions, with "undetermined coefficients" (generic A,B,C..etc) So if the right side of your differential equation has say "sin(3x)", you would chose: Acos(3x)+Bsin(3x). Because if you differentiate sin(3x) and cos(3x) over and over, you always get sin(3x) and cos(3x) off by only a scalar multiple (this "coefficient").
Think about it, if you have a function y=e^x and you have a differential equation, this means you've differentiated e^x right? Which means you would expect there to be "e^x"s. This guides your "guess". I can explain as I type it out if that doesn't make sense.
Yes, understood. Browsed on google and found explanation in "pauls online notes" site.
Good, that is exactly what I was going to link you. So to form out particular solution we look at the right side of our equation and we see: \[...=5x-\cos(2x)e^{-x}\] So, to form our particular solution, we are going to "guess" something in the form: \[y_P(x)=Ax+B+e^{-x}(D \cos(2x)+E \sin(2x))\] The Ax+B comes from the 5x, because 5x is a first order polynomial where the constant term is zero. The e^(-x) DOES NOT have a constant BECAUSE I have let the D and E absorb it respectively (ignoring C because Ccos(2x) bothers me). And the cos(2x)+... comes from the cos(2x) because if you differentiate cos(2x) and sin(2x) you get them over and over off by a constant.
Follow me so far?
yes.
Okay, so now all we need to do is differentiate it once, differentiate it twice, a third time, and a forth, and plug it into the above equation^^^ and set it equal to =5x-cos(2x)e^(-x) and solve the constants (for example, if we have Ax+B=2x+3, we can deduce A=2 and B=3).
I guess we should just start differentiating? I'll do it in wolfram and post it...
This is the first one according to wolfram: \[y_P'(x)=A-2 D e^{-x} \sin(2 x)-D e^{-x}\cos(2 x)-Ee^{-x}\sin(2 x)+2 Ee^{-x} \cos(2 x)\]
Ohh. The forth one will be long
And so on...I guess haha. Anyways, you plug that into the original and you would get: \[y_P^{(4)}(x)+3y_P'''(x)+2y_P''(x)=5x-\cos(2x)e^{-x}\] Then set ALL the coefficients of the x terms equal to 5, all the coefficients of cos(2x)e^(-x) equal to -1 and the coefficients of the rest to 0. Its hard to explain but does that make sense? A better example would be looking at paul notes :/
I'll write the whole thing down fully and if I have any trouble, I will say. Big thanks for now.
Okay, no problem.
just by the look of it, I'm guessing, there wont be any x on the left side of that equation, as none of the derivatives contains a simple x.
Yeah
So, if I hadn't made any stupid mistakes, I get \[e^{-x}(-10Dsin(2x)+20Dcos(2x)+20Esin(2x)+10Ecos(2x))=5x-e^{-x}(\cos2x)\]
the right side is the original one, it just didn't fit I guess.
Okay, so that would mean that -cos(2x)e^(-x), coefficient -1, would have to be equal to the coefficient of 20D+10E=-1, -10D+20E=0. There may be an x^4 term so the 4th derivative will give the linear term.
"There may be an x^4 term so the 4th derivative will give the linear term." didn't get this one.
E=-1/50 D=-1/25
I'm a bit confused about the x's.
I'm no expert, but maybe we needed that partial solution to contain x^2 or x^3 or something like that?
You have to differentiate 4 times. So if an x^4 term was added by the time you differentiated it 4 times you would get an x term. Which you could find the coefficient for. Do the calculation but I think it will be: \[\frac{5}{4!}=\frac{5}{24}\] Just doing mental stuff lol
you mean just instead of Ax I put Ax^4 ?
Yes. I believe.
So taking only the x term part (cos and sin part is done I believe): \[A*x^4+B\] \[(1) 4Ax^3\]\[(2) 12Ax^2\]\[(3) 24Ax\]\[(4) 24A\] Where (1)... are differentials. That leaves us with\[24A+72Ax+24Ax^2=5x\]and that doesn't make any sense, because here A must be 0 and 5/72. I think, we need some kind of x term at B.
Try, Ax^4+Bx^3+Cx^2+Dx+E?
Look at this: http://www.wolframalpha.com/input/?i=solve+y''''(t)%2B3y'''(t)%2B2y''(t)%3D5t-cos(2t)e%5E(-t)
OHHHH!!!!! THE HOMOGENOUS SOLUTION HAS AN X IN IT!!!! SO THERE IS A REDUNDENCY! Therefore you have to multiply by another x^2! That makes the term an x^3! That should fix it I think?
So try x^2(Ax+B)+(rest)
Yeah, thanks, solved it. Just curious, could you explain the redundancy in few words?
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