# #2 Variation of parameters with complemetry EQ (Diffy Q)

1. Apr 9, 2006

### Weatherkid11

Given that http://forums.cramster.com/Answer-Board/Image/cramster-equation-200641003458632802260985487500893.gif [Broken] is the complementary function for the differential equation http://forums.cramster.com/Answer-Board/Image/cramster-equation-200641003635632802261951581250765.gif [Broken] use the variation of parameters method to find a particular solution of yp
Well i think i use the wronskian 1st, so that would be w= http://forums.cramster.com/Answer-Board/Image/cramster-equation-2006410038406328022632042375002059.gif [Broken] then i use the integrals to find u1 and u2. But what do I do next?

Last edited by a moderator: May 2, 2017
2. Apr 10, 2006

### HallsofIvy

Anothere one? You seem to have imperfectly memorized formulas rather than actually learning what to do. Where did you get the formulas to memorize? Don't you have a text book? This is a good example of why it is better to LEARN concepts than to memorize formulas.

Since you know that x2 and x3 satisfy the homogeneous equation, you look for solutions to the entire equation of the form y(x)= u(x)x2+ v(x)x3. Then y'= u'x2+ 2ux+ v'x3+ 3vx2. Now, require that
u'x2+ v'x3= 0. That leaves y'= 2ux+ 3vx2 and y"= 2u'x+ 2u+ 3v'x2+ 6vx. Putting that into the differential equation, x2y"- 4xy'+ 6y= x2(2u'x+ 2u+ 3v'x2+ 6vx)- 4x(2ux+ 3vx2)+ 6(ux2+ vx3= 2x3u'+ (2- 8+ 6)x2u+ 3x4v'+ (6- 12+ 6)x3u= 2x3u'+ 3x4v'= x3. For x not 0 (which is a singular point anyway) 2u'+ 3xv'= 1. Because we required that u'x2+ v'x3= 0, there is no u" or v" and because x2 and x3 satisfy the homogeneous equation, all terms involving u and v without derivative cancel so we have only an equation in u' and v'.
We also have the requirement u'x2+ v'x3= 0 or, (again for x not 0), u'+ xv'= 0.

Solve the two equations u'+ xv'= 0 and 2u'+ 3xv'= 1 algebraically for u' and v' and then integrate.