*This preview shows
pages
1–3. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **MATH 535: Introduction to Partial Differential Equations Spring11 (F.J. Sayas) Some differential equations (Refresher for Section 1.2) Linear differential equations of order one The general form of a linear differential equation of order one is y + a ( x ) y = f ( x ) . To solve it, we start by finding an antiderivative of a ( x ) = Z a ( x )d x. We multiply both sides of the equation by e to get ( e ( x ) y ) = e ( x ) ( y + a ( x ) y ) = e ( x ) f ( x ) . Finally, we look for an antiderivative of e f G = Z e ( x ) f ( x )d x to obtain e ( x ) y = G ( x ) + c, that is, y = ce- ( x ) + e- ( x ) G ( x ) . Therefore, everything is reduced to multiplying the equation by e R a ( x )d x and then integrating both sides of the equation. 2. Equations with separable variables These are equations that can be written in the form y = g ( y ) f ( x ) . To solve them we need to find respective antiderivatives of 1 /g and f H = Z 1 g ( y ) d y F = Z f ( x )d x. Then H ( y ) = F + c, with c R is an implicit form of the set of solutions to the equation. A way to remember this process is the following one. Rewrite the equation d y d x = y = g ( y ) f ( x ) as d y g ( y ) = f ( x )d x and integrate both sides Z d y g ( y ) = Z f ( x )d x + c. 1 Some annoying integrals All the following antiderivatives can be computed using integration by part (some times applied several times) or with the following procedure. I am just giving an example. General rules can be easily derived. To compute Z xe x d x, note that this is a particular solution of the equation y = xe x . We can find this with indeterminate coefficients, that is, looking for a solution of the form y = ( Ax + B ) e x A and B are constants to be determined) and then taking the derivative in both sides, equating xe x = y = ( Ax + B ) e x + Ae x to easily determine A and B x = Ax + B + A ( A = 1 B + A = 0 . This process can be applied to compute some of the following integrals: Z x k e x d x = ( A k x k + A k- 1 x k- 1 + ... + A ) e x Z x cos( x )d x = ( Ax + B )cos( x ) + ( C x + D )sin( x ) Z x sin( x )d x = ( Ax + B )cos( x ) + ( C x + D )sin( x...

View
Full
Document