variation of parameters

the undetermined coefficient method may not prove out to be useful all the time ..we may need to follow a little different path...

equation we have is Y''+const.Y'+Y = G(x)

assume particular solution is Yp = u1y1+u2y2 ............(1)

Yp' = u1'y1+u1y1'+u2'y2+u2y2'

assume u1'y1 + u2'y2' = 0 then Yp' = u1y1'+u2y2'

Yp'' = u1'y1'+u2'y2'+u1y1''+u2y2''

putting all these values in (1)...we have

u1'y1'+u2'y2'+u1y1''+u2y2''+const. * u1y1'+const.u2y2'+u1y1+u2y2 = G(x)

u1(y1''+const.y1'+y1)+u2(y2''+const.y2'+y2)+u1'y1'+u2'y2' = G(x)

y1''+const.y1'+y1 = 0 as per the general solution ..we are left with

u1'y1'+u2'y2' = G(x) ........(2)

u1'y1+u2'y2 = 0 ...(3)

u1' = -u2'y2/y1

u2' = G(x)*y1 / (y2'y1-y2y1')..............(4)

u1' = -G(x)*y2 / (y2'y1-y2y1').............(5).

substituting 4 and 5 in (1)...we have

Yp = y1*integration(-G(x)*y2 / (y2'y1-y2y1'))+y2*integration(G(x)*y1 / (y2'y1-y2y1'))

Undetermined method of coefficients

in a non homogeneous equation, EQUATION will look like of the form below -

Y''+Y'*const.+Y = G(x) ................0

assume equation has solution Y1 and Y2...

however equation Y''+Y'*const.+Y = 0 ...............1

we know (1) has a solution c1*y1+c2*y2......

now putting Y1-Y2 in equation 1 will give us

(Y1-Y2)''+(Y1-Y2)'*const.+(Y1-Y2) = 0

Y1''+const.*Y1'+Y1 - Y2''-Y2'*const.-Y2 = 0

G(x)-G(x) =0 ..hence proved Y1-Y2 i solution to (1)

and we can safely say

Y1-Y2 = c1y1+c2y2 where Y1 and Y2 can be any solution to our non homogeneous equation (0)

we can r.h.s of equation (2) as complementary solution while Y2 as particular solution and get value of Y1 using initial / boundary values

Y''-5Y'+6Y = sin(4x) ...........................3 Y(0) = 10 Y'(0)= 15

complementary solution will be --- c1*exp(2x) + c2*exp(3x)

let us take particular solution to be --------c3 * sin4x + c4* cos4x

put particular solution in 3 ..we have

-16*c3 sin4x - 16c4cos4x -20c3cos4x + 20 c4sin4x+6c3sin4x + 6c4cos4x = sin4x

equating both the side..

-16c3+20c4+6c3= 1 and -16c4-20c3+6c4 = 0

20c4-10c3 = 1 and 20c3- 10c4 = 0

c3 = 1/30 c4 = 1/15

particular solution then is 1/30 sin4x+1/15 cos4x

Y1 = c1*exp(2x) + c2*exp(3x) + 1/30 sin4x + 1/15 cos4x ...

x= 0 Y =0

c1 + c2 + 1/15 = 0... 4 (i)

Y'(0) = 0

2c1+ 3c2+ 4/30 = 0 ..4(ii)

c1 = -1/15

and solution is

Y1 = -1/15*exp(2x)+1/30 sin4x+1/15 cos4x