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Finding a particular solution (Diff. Eq.) 
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#1
Jul2807, 09:36 AM

P: 109

1. The problem statement, all variables and given/known data
I seem to not fully understand how to find particular solutions. I'm having a hard time guessing what the solutions maybe. I'll explain as I write out the problem. [tex]y'''  2y'' + y' = te^t + t^2 + 4[/tex], find the general solution. 2. Relevant equations [tex]Z(r) = r^3  2r^2 + r[/tex] 3. The attempt at a solution Found the homogenous solution: [tex]r^3  2r^2 + r = 0[/tex] [tex]r (r  1)^2 = 0[/tex] [tex]r_1 = 0, \ r_2 = 1, \ r_3 = 1[/tex] [tex]y = C_1 + C_2e^t + C_3te^t[/tex] Now I'm trying to find the particular solutions. For [tex]te^t[/tex], I assume: [tex]y_1 = At^2e^t[/tex] [tex]y_1' = Ae^t (t^2 + 2t)[/tex] [tex]y_1'' = Ae^t (t^2 + 4t + 2)[/tex] [tex]y_1''' = Ae^t (t^2 + 6t + 6)[/tex] Now I plug back in: [tex]Ae^t (t^2 + 6t + 6) 2Ae^t (t^2 + 4t + 2) + Ae^t (t^2 + 2t) = te^t[/tex] [tex]2A = t[/tex] [tex]A = \frac{t}{2}[/tex] [tex]y_1 = \frac{1}{2}t^3e^t[/tex] Now, for [tex]t^2[/tex], I assumed [tex]y_2 = Bt^3[/tex], but in an example in the book that had [tex]t^2[/tex], it said that [tex]y_2 = t^2 (A_1t^2 + A_2t + A_3)[/tex]. I don't understand how they got that. Also, for the constant 4, I assumed [tex]y_3 = Ct[/tex]. Is this correct? I don't fully understand how I am supposed to go about guessing the solutions. There must be some guidelines, but we weren't taught what they are. If someone could help me with this, it would be greatly appreciated. 


#2
Jul2807, 11:21 AM

P: 1,875

Are you trying to use superposition to get solutions?
Generally, it works out that you guess the type of function that's the same as the particular part. So if you have ODE = k, where ODE is some kind of diff eq, then you should guess a constant as the solution. Another example would be if you had ODE = cos(2t) you should guess Acos(2t) + Bsin(2t) or simply exp(i2t). Sometimes you will have a mixture of functions, in which case your guess should also be a mixture. Say there was a polynomial with [tex]ODE = At^2 + Bt + C[/tex], you should just a polynomial of the same form [tex]y_p = \alpha t^2 + \beta t + \gamma[/tex] If you had something of a mixture, such as [tex]ODE = e^{\alpha t} P_2(t)[/tex] you should guess [tex]y_p = exp(\alpha t)(At^2 + Bt + C)[/tex] for each case just extend the generalization further and further [tex]ODE = e^{\alpha t} P_2(t) sin(\beta t)[/tex] gives [tex]y_p = exp(\alpha t)sin(\beta t)(At^2 + Bt +C) + exp(\alpha t)}cos(\beta t)(At^2 + Bt + C)[/tex] 


#3
Jul2807, 11:32 AM

P: 109

But don't I have to look back to see if the solution I'm guessing already exists?
For example, if I have: [tex]ODE = e^t[/tex] My homogenous solution comes out to be: [tex]y = C_1e^t + C_2te^t[/tex] and I need to guess a solution for [tex]e^t[/tex], I was told that I can't guess that [tex]y_1 = Ae^t[/tex], because [tex]e^t[/tex] is already a solution. Neither can I choose [tex]y_1 = Ate^t[/tex], because it's also a solution. So I need to use [tex]y_1 = At^2e^t[/tex]. So do I just need to look back to make sure that what I'm guessing is not a solution already? 


#4
Jul2807, 04:49 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310

Finding a particular solution (Diff. Eq.)
[tex]y_1 = \frac{1}{2}t^3e^t[/tex] Now, for [tex]t^2[/tex], I assumed [tex]y_2 = Bt^3[/tex], but in an example in the book that had [tex]t^2[/tex], it said that [tex]y_2 = t^2 (A_1t^2 + A_2t + A_3)[/tex]. I don't understand how they got that. Also, for the constant 4, I assumed [tex]y_3 = Ct[/tex]. Is this correct? I don't fully understand how I am supposed to go about guessing the solutions. There must be some guidelines, but we weren't taught what they are. If someone could help me with this, it would be greatly appreciated.[/QUOTE] Have you looked in your textbook? I don't know of any that doesn't go into pretty specific detail about how to "guess" solutions. The kind of solutions that you expect for any "linear homogenous d.e. with constant coefficients: are: 1) exponentials: Ce^{kx} 2) sine or cosine: Ccos(kx) or Csin(kx) 3) polynomials: Cx^{n}+ Dx^{n1}+ ...+ Yx+ Z. 4) Combinations of those: (Ax^{n}+ ...)e^{kx}(Ucos(mx)+ Dsin(nx) The method of "undetermined coefficients" only works if the righthandside of the equation is one of those forms. In general, if it is, you try that same type. That is, if the rhs is e^{kx} you try Ae^{kx}. If the rhs is either sin(kx) or cos(kx), you try Asin(kx)+ Bcos(kx) (in general you will need both sin(kx) and cos(kx)). If the rhs is a polynomial or a polynomial times an exponential you try the same thing: however, if the highest power of x is n, you may need all powers from n on down even if the lower powers do not all appear in the rhs. The reason Ae^{t} doesn't work here is that e^{t} is that it, as well as te^{t} and t^{2}e^{t} are already solutions to the homogeneous equation. You need to try At^{3}e^{t}. 


#5
Jul2907, 09:34 AM

P: 109

[tex]\begin{array}{lcr} g_i(t) & Y_i(t)\\\hline \\ P_n(t) = a_0t^n + a_1t^{n1}+ ... + a_n & t^s(A_0t^n + A_1t^{n1} + ... + A_n)\\\\P_n(t)e^{\alpha t} & t^s(A_0t^n + A_1t^{n1} + ... + A_n)e^{\alpha t}\\\\P_n(t)e^{\alpha t}\left\{\begin{array}{cc} \sin{\beta t\\ \cos{\beta t}\end{array} \right. & t^s[(A_0t^n + A_1t^{n1} + ... + A_n)e^{\alpha t} \cos{\beta t} + (B_0t^n + B_1t^{n1} + ... + B_n)e^{\alpha t} \sin{\beta t}]\end{array}[/tex] So I understand how to choose them, but I wasn't sure whether I need to look and see if what I chose is already a solution. Now I get it, Thanks Mindscrape and HallsofIvy. 


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