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

[Weatherkid11]

Given that http://forums.cramster.com/Answer-Board/Image/cramster-equation-200641003458632802260985487500893.gif is the complementary function for the differential equation http://forums.cramster.com/Answer-Board/Image/cramster-equation-200641003635632802261951581250765.gif 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 then i use the integrals to find u1 and u2. But what do I do next?

 

[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 textbook? 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.