Ordinary differential equations

[Telemachus]

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
y''+P(x)y'+Q(x)y=0 (1)

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

y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx
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 w(y_1,y_2,x) must be zero. So, I just made the calculus for the wronskian:

\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--SOLVED--

 

[LCKurtz]

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
y''+P(x)y'+Q(x)y=0 (1)

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

y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx
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 w(y_1,y_2,x) must be zero.

You mean if they are linearly dependent, W = 0.

So, I just made the calculus for the wronskian:

\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

Looks fine.

 

[Telemachus]

Right. Thanks.