# Thread: Solve the ordinary diferential equation

1. ## Solve the ordinary diferential equation

Solve the ordinary differential equation
x2y'' - 3xy' + 3y = x3cosx
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)Ckxk, and also for y' and y'' into the equation. also for the RHS too.

2. ## Re: Solve the ordinary diferential equation

First, find two solutions for x²y'' - 3xy' + 3y =0
Let y=x^r

3. ## Re: Solve the ordinary diferential equation

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?

4. ## Re: Solve the ordinary diferential equation

So, what are the two particular solutions y(x) that you found ?
As a consequence, what is the genertal solution (i.e. all the solutions) of x²y''-3xy'+3y=0 ?

5. ## Re: Solve the ordinary diferential equation

I'm not sure if it's based on x or r. If it's based on r, the solution would be: C1e^x + C2e^(3x). sorry I'm just so confused since there're so many methods...

6. ## Re: Solve the ordinary diferential equation

All right, the general solution of x²y'' - 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 x²y'' - 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" ?

7. ## Re: Solve the ordinary diferential equation

Hello. Thank you so much for that. I've solved the question.

8. ## Re: Solve the ordinary diferential equation

Originally Posted by JJacquelin
All right, the general solution of x²y'' - 3xy' + 3y =0 is y=C1e^x + C2e^(3x).
NO, it's not- and I am sure you know better! Was this just an "extended typo"? The "characteristic equation" tran got by setting $y= x^r$ was $r^2- 4r+ 3= 0$ which has roots r= 1 and r= 3 so the general solution is $y= C_1x+ C_2x^3$ as you say below.

(The substitution t= ln(x) or $x= e^t$ converts this to a differential equation with constant coefficients with the same characteristic equation so solution $y= C_1e^t+ C_2e^3t= C_1e^{ln(x)}+ C_2e^{3ln(x)}= C_1e^{ln(x)}+ C_2e^{3ln((x)}= C_1x+ C_2x^3$.)

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 x²y'' - 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" ?

9. ## Re: Solve the ordinary diferential equation

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.

10. ## Re: Solve the ordinary diferential equation

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 $y= \sum_{k=0}^\infty C_kx^k$ so that $y'= \sum_{k=1}^\infty kC_n x^{k- 1}$ and $y''= \sum_{k= 2} k(k-1)x^{k-2}$. Putting those into the equation,
$\sum_{k= 2}^\infty C_kk(k-1)x^k- 3\sum_{k= 1}^\infty C_k x^k+ 3\sum_{k=0}^\infty C_kx^k= x^3(\sum_{k= 0}^\infty \frac{(-1)^n}{(2k+1)!}x^{2k+ 1}$.

Now look at individual values of the power of x. We have $x^0$ only in the third sum on the left and NOT on the right: $C_0= 0$. We have $x^1$ in the seconds and third sums on the left but not on the right: $-3C_1+ 3C_k= 0$ for any $C_k$: That gives the " $C_1x$" above. For $x^2$ or higher, we have all three terms on the left.
$\sum_{k=2}^\infty (k(k- 1)- 3k+ 3)C_k= \sum_{k= 3}^\infty \frac{(-1)^n}{(2n+1)!}x^{2k+4}$.

When k= 2, that is $(2(1)- 3(2)+ 3)C_1= (-1)C_2= 0$ so that $C_2= 0$. When k= 3, [tex](3(2)- 3(3)+ 3)C_3= 0C_3= 0[tex] which is true for any $C_3$- that is where the " $C_2x^3$" above comes from.

### solution to 3xy'' 3y' xÂ²y = 0

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