When performing variation of parameters to find the particular solution of a differential equation, I am confused whether the integrals should be definite or indefinite.

Consider the differential equation

$\frac{d^2 y}{dx^2} = f(x)$

integrate twice

$y = c_2 + c_1x + \int\int f(x) \hspace{30 mm} (1)$

Compare this to using variation of parameters to find y.

The homogeneous solution is

$y_h = c_1x+c_2$

which makes

$y_1 = x$

$y_2 = 1$

The Wronskian is

$W = -1$

Using variation of parameters with indefinite integrals, the particular solution is

$y_p = -y_1\int \frac{y_2 f(x)}{W}dx + y_2\int \frac{y_1 f(x)}{W}dx$

for this case

$y_p = x\int f(x)dx + \int x f(x)dx$

Perform integration by parts on the last term and

$y_p = x\int f(x)dx - x\int f(x)dx + \int\int f(x)$

Remove the cancelling terms

$y_p = \int\int f(x)$

$y = y_h + y_p$

$y = c_2 + c_1x + \int\int f(x) \hspace{30 mm} (2)$

The same as $(1)$. However, I have seen that definite integrals can be used with variation of parameters when $f(x)$ is complex and cannot be integrated analytically. This would make the particular solution

$y_p = -y_1\int_0^x \frac{y_2 f(x)}{W}dx + y_2\int_0^x \frac{y_1 f(x)}{W}dx$

which ends up becoming

$y = c_2 + c_1x + \int_0^x\int_0^s f(t)dt ds \hspace{30 mm} (3)$

which doesn't make sense to me because if we go back to just integrating the original equation twice it would be like integrating with definite integrals but still including the constants. Can someone explain where my thinking is in error?



1 Answer 1


More precisely

$$\int \int f(x)\,dx\,dx = c_2 + c_1x + \int_0^x \int_0^s f(x)\,dt\,ds$$

where $c_1 = f'(0)$ and $c_2 = f(0)$

This is why I find the anti-derivative notation to be sometimes confusing.

  • $\begingroup$ I agree the anti-derivative notation can be confusing. Do you know any references that explain it well? $\endgroup$
    – T Mac
    Apr 13, 2015 at 15:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .