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Ordinary differential equations

  • Thread starter Telemachus
  • Start date
832
30
1. Homework Statement
Hi there. Well, I have some doubts about this exercise, I think I've solved it, but I wanted your opinion, which always help. So, it says:

There is no general method for solving the homogeneous general equation of second order
[tex]y''+P(x)y'+Q(x)y=0[/tex] (1)

But if we already know a solution [tex]y_1(x)\neq 0[/tex], then we always can find a second solution linearly independent

[tex]y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx[/tex]
Demonstrate that this functions form a basis for the space of solutions of the differential equation (1).


So, what I did is simple. I know that if the solution is linearly independent, then the Wronskian [tex]w(y_1,y_2,x)[/tex] must be zero. So, I just made the calculus for the wronskian:

[tex]\left| \begin{matrix}{y_1}&{y_2}\\{y_1'}&{y_2'}\end{matrix} \right|=\left| \begin{matrix}{y_1}&{y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx}\\{y_1'}&{y_1'(x) \int \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2}dx+y_1(x) \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2(x)}dx}\end{matrix} \right|=e^{- \int P(x)dx}\neq0[/tex]


--SOLVED--
 
Last edited:

LCKurtz

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1. Homework Statement
Hi there. Well, I have some doubts about this exercise, I think I've solved it, but I wanted your opinion, which always help. So, it says:

There is no general method for solving the homogeneous general equation of second order
[tex]y''+P(x)y'+Q(x)y=0[/tex] (1)

But if we already know a solution [tex]y_1(x)\neq 0[/tex], then we always can find a second solution linearly independent

[tex]y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx[/tex]
Demonstrate that this functions form a basis for the space of solutions of the differential equation (1).


So, what I did is simple. I know that if the solution is linearly independent, then the Wronskian [tex]w(y_1,y_2,x)[/tex] must be zero.
You mean if they are linearly dependent, W = 0.

So, I just made the calculus for the wronskian:

[tex]\left| \begin{matrix}{y_1}&{y_2}\\{y_1'}&{y_2'}\end{matrix} \right|=\left| \begin{matrix}{y_1}&{y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx}\\{y_1'}&{y_1'(x) \int \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2}dx+y_1(x) \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2(x)}dx}\end{matrix} \right|=e^{- \int P(x)dx}\neq0[/tex]
Looks fine.
 
832
30
Right. Thanks.
 

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