First, find two solutions for x球'' - 3xy' + 3y =0
Let y=x^r
Solve the ordinary differential equation
x^{2}y'' - 3xy' + 3y = x^{3}cosx
I am working on this question for 2 days and I coudn't get to the final result.
we have to use the power series to solve this question. so I substitute y= sum(n=0 to infi)Ckx^{k}, and also for y' and y'' into the equation. also for the RHS too.
please help me to solve it.
Thank you so much for the suggestion. but I don't really get it. if you let y= x^r then y' = rx^(r-1), y'' = r(r-1)x^(r-2)
then substitute y,y',y'' in the equation. I got r=3,r=1. then what's the next step to solve this equation?
All right, the general solution of x球'' - 3xy' + 3y =0 is y=C1e^x + C2e^(3x).
But what do you mean : << if it's based on x or r >> ? Remember : You let y=x^r and you found r=1 and r=3. Hense y= x^1 and y=x^3 are two solutions. What is confusing ?
Then, come back to the complete equation x球'' - 3xy' + 3y = (x^3)cosx
Now, you have to find a particular solution for the complete equation. Only one will be sufficient ! But it is the most difficult part of the job. Did you yeard about "the variation of the constant" ?
NO, it's not- and I am sure you know better! Was this just an "extended typo"? The "characteristic equation" tran got by setting was which has roots r= 1 and r= 3 so the general solution is as you say below.
(The substitution t= ln(x) or converts this to a differential equation with constant coefficients with the same characteristic equation so solution .)
But what do you mean : << if it's based on x or r >> ? Remember : You let y=x^r and you found r=1 and r=3. Hense y= x^1 and y=x^3 are two solutions. What is confusing ?
Then, come back to the complete equation x球'' - 3xy' + 3y = (x^3)cosx
Now, you have to find a particular solution for the complete equation. Only one will be sufficient ! But it is the most difficult part of the job. Did you yeard about "the variation of the constant" ?
in this case r=3, r=1. => y1= x^3, y2=x
then we need to divide throughout by x^2 to get the standard form. After that we need to find the Wronskian W(y1,y2) which is equal to -2x^3.
then find u1, u2 by integrating u1'(x), u2'(x). the general sol is y = u1y1+ u2y2.
I was too late to edit the above but:
Also the original post said "we have to use the power series to solve this question. so I substitute y= sum(n=0 to infi)C_kx^k" so this method isn't appropriate. What is appropriate is to write so that and . Putting those into the equation,
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Now look at individual values of the power of x. We have only in the third sum on the left and NOT on the right: . We have in the seconds and third sums on the left but not on the right: for any : That gives the " " above. For or higher, we have all three terms on the left.
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When k= 2, that is so that . When k= 3, [tex](3(2)- 3(3)+ 3)C_3= 0C_3= 0[tex] which is true for any - that is where the " " above comes from.