0
$\begingroup$

Hey everyone I need some help understanding how to use Lagrange method and specifically the Wronskian used in it to solve 2nd order non homogeneous ode. Let :

$$y''(t)+a_1y'(t)+a_2y(t)=f(t)$$

I know that the general solution to a non homogeneous 2nd order ode is:

$$y(t)=c_1y_1(t)+c_2y_2(t)+y_p(t)$$

Where $y_1$ and $y_2$ are two linear independant solutions of the homogenous ode and $y_p$ is a particular solution of the non homogeneous ode.

I also know that we can find $y_p$ using the Wronskian in the formula:

$$y_p=\int_{t_0}^t \frac{y_1(s)y_2(t)-y_1(t)y_2(s)}{W_{(y_1,y_2)}(s)}f(s)ds$$

Here's where I'm confused:

I'm not sure what $t_0$ should be... I think that any $t_0 \in I$, where $I$ is the interval for which the two solutions $y_1$ and $y_2$ are continuous will do but I'm still not sure how to choose the value to substitute to $t_0$.

For example for the ode: $y''-y'-2y=e^{-t}$ I found that two linear independant solutions of the homogeneous are $y_1=e^{-t}$ and $y_2=e^{2t}$ and the particular solution should be:

$y_p=\int_{t_0}^t \frac{y_1(s)y_2(t)-y_1(t)y_2(s)}{W_{(y_1,y_2)}(s)}f(s)ds$

The Wronskian is $W(s)=3e^s $ so

$y_p=\int_{t_0}^t \frac{e^{-s}e^{2t}-e^{-t}e^{2s}}{3e^s}e^{-s}ds$

I could go on to integrate the function but I'm not sure what I should substitute $t_0$ with.. My books solution suggests $t_0=0$ but why?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ Be carefull at last line it should be $$y_p=\int_{t_0}^t \frac{e^{-s}e^{2t}-e^{-t}e^{2s}}{3e^s}\color{red}{e^{-s}}ds$$ It's a function of s not a function of t $\endgroup$ – Aryadeva Jan 23 at 17:06
1
$\begingroup$

$$y_p=\int_{t_0}^t \frac{e^{-s}e^{2t}-e^{-t}e^{2s}}{3e^s}e^{-s}ds$$ $$y_p=\frac 13e^{2t}\int_{t_0}^t {e^{-s}ds-\frac 13e^{-t}\int_{t_0}^t }ds$$ At $t_0$ the function in the integrand becomes a constant. At $t$ you get a function of t. I don't know if I am clear. You should evaluate the integral at $t_0=0$ as suggested by your book and also at different values to see what happens. You end with the same answer because the homogeneous solution will absorb the extra terms of the particular solution.

$$y_p=\frac 13e^{2t}\int_{t_0}^t {e^{-s}ds-\frac 13e^{-t}\int_{t_0}^t }ds$$ $$y_p=-\frac 13e^{2t}|e^{-s}|_{t_0}^t -\frac 13e^{-t}|s|_{t_0}^t $$ $$y_p=-\frac 13e^{2t}(e^{-t}-e^{t_0})-\frac 13e^{-t}(t-{t_0}) $$ $$y_p=-\frac 13e^{2t}(e^{-t}-e^{t_0})-\frac 13e^{-t}(t-{t_0}) $$ The first term is absorbed byt the homogeneous solution. It remains only the following term: $$\boxed {y_p=-\frac 13e^{-t}t}$$ The rest are extra terms as you can check...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So if the value of $t_0$ doesn't matter why not use indefinite integral? $\endgroup$ – SlimJim Jan 23 at 16:00
  • $\begingroup$ Of course you can use indefinite integral too @SlimJim $\endgroup$ – Aryadeva Jan 23 at 16:08
  • $\begingroup$ Well sorry if I seem ignorant but is there any advantage in using definite integral? $\endgroup$ – SlimJim Jan 23 at 16:42
  • 1
    $\begingroup$ We are all ignorant.. Well I don't see any advantage for the particular solution to use definite integral. But maube in some physics application it may be usefull. In this case it changes nothing. @SlimJim $\endgroup$ – Aryadeva Jan 23 at 16:45
  • $\begingroup$ You made a mistake in your last formula....That's chy i couldnt find the answer....It should be $f(s)$ but you wrote $f(t)$ @SlimJim $\endgroup$ – Aryadeva Jan 23 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.