I am reading in my textbook about the derivation of the general solution for $$\dfrac{dy}{dx} + P(x)y=f(x)$$

First, the textbook shows that a general solution for this DE is composed by the sum of a particular solution of the DE, $y_p$, and the general solution, $y_c$, of the associated homogeneous equation $\dfrac{dy}{dx} + P(x)y=0$. That is, $$y=y_c+y_p$$

The book proceeds to use separation of variables to conclude $y_c=ce^{-\int P(x)dx}$, and the text assigns $y_1(x)= e^{-\int P(x)dx}$ so that $y_c=cy_1(x)$

Then the book proceeds:

"We can now find a particular solution of equation (2) [the original, non-homogeneous linear ODE] by a procedure known as variation of parameters. The basic idea here is to find a function $u$ so that $y_p=u(x)y_1(x)$....In other words, our assumption for $y_p$ is the same as $y_c=cy_1(x)$ except that $c$ is replaced by the 'variable parameter' $u$."

The book makes it seem that variation of parameters is a common and well established technique for solving DE. Is there a solid reason of why variation of parameters was used, or is it just a technique that we try out, and sometimes works and sometimes doesn't?

Why does the book say "In other words, our assumption for $y_p$ is the same as $y_c=cy_1(x)$"? There always exists such a $u$, which can be expressed as $u(x)=\dfrac{y_p(x)}{y_1(x)}$ (the denominator will never be zero). I don't see the need for using the work "assumption" since I am not sure that we are taking anything for granted. Is this just poor wording for "we set $y_p=u(x)y_1(x)$, which has a similar form as $y_c=cy_1(x)$"?

  • $\begingroup$ I think the book intended the same thing as your modified sentence "we set $y_p=u(x)y_1(x)$, which has similar form as..." Indeed we always can write $y_p$ in that form, but it is still useful to do so. In more complex multi-variable PDE problems you often "assume" a solution $f(x,y)$ has a form $f(x,y)=h(x)g(y)$ (which is not a general form), and then show that form actually works. $\endgroup$ – Michael Jan 25 '16 at 17:42
  • $\begingroup$ @Michael Is there something that hints us to use variation of parameters, or is it just works or doesn't work sometimes? For example, did the fact that $y=y_c+y_p$ indicate that this problem has characteristics which hint that it is solvable by variation of parameters? $\endgroup$ – Ovi Jan 25 '16 at 19:02

The intuition behind $y_p(x) = u(x)y_1(x)$ is that when we differentiate using the product rule, we already know something nice about the derivative of $y_1(x)$. It seems to work well in this linear first-order case (which is a large class of problems). I don't know if it will work out for other problems, let's try it for second order linear problems:

Inhomogeneous ODE: $y''(x) + P(x)y'(x) + Q(x)y(x) = f(x) $

Homogeneous version: $y''(x) + P(x)y'(x) + Q(x)y(x) = 0$

Let $y_1(x)$ be a nonzero solution to the homogeneous equation. Let's try the form:
$$ \boxed{y(x) = u(x)y_1(x)}$$

Differentiating gives: \begin{align} y(x) &= u(x)y_1(x)\\ y'(x) &= u(x)y_1'(x) + u'(x)y_1(x)\\ y''(x) &= u(x)y_1''(x) + u'(x)y_1'(x) + u''(x)y_1(x) + u'(x)y_1'(x) \end{align} Thus: \begin{align} y''(x) &= u(x)y_1''(x) + 2u'(x)y_1'(x) + u''(x)y_1(x)\\ P(x)y'(x) &= P(x)u(x)y_1'(x) + P(x)u'(x)y_1(x)\\ Q(x)y(x) &= Q(x)u(x)y_1(x) \end{align} Adding the above three equations, setting the result to $f(x)$, and using the fact that $y_1''+Py_1'+Qy_1=0$ gives: $$ f(x) = y''(x) + P(x)y'(x) + Q(x)y(x) = u''(x)y_1(x) + u'(x)(2y_1'(x)+P(x)y_1(x)) $$ Well now this is pretty nice because we can define $g(x) = u'(x)$ and we get: $$ f(x)/y_1(x) = g'(x) + g(x)\left(\frac{2y_1'(x) + P(x)y_1(x)}{y_1(x)}\right) $$ This is in fact a linear (inhomogeneous) first order equation in $g(x)$, so we can solve it. Thus: The technique is also useful here!

  • $\begingroup$ In general, I think this technique can reduce a linear $n$th order inhomegeneous ODE to a linear $(n-1)$th order inhomogeneous ODE. $\endgroup$ – Michael Jan 25 '16 at 19:39

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.