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

Higher order ODE

Higher order ODE are little tricky to solve and can be classified into homogeneous and non homogeneous.

a normal OD will be of the format below -

Y'''+const.Y''+const.Y'+Y = G(X)

when G(X) = 0 equation is called homogeneous otherwise when G(X) not equal 0 then it is non homogeneous.

Basic method to solve both the equations is same however change will be introduced depending upon type of equation.

Let us first look at homogeneous solution-

we have equation -

Y''+const.Y'+Y = 0 .......................1

assume Y = exp(rt)

then Y' = r*exp(rt)

Y'' = r*r*exp(rt)

using this equation (1) reduces to

exp(rt) (r*r + const.*r+1) = 0 .................................2

exp(rt) can't be zero

r*r+const.*r+1 = 0 can be solved for roots of the equation

Solution of ODE will look like -

Y = c1*exp(r1*t) + c2*exp(r2*t) where r1 and r2 are roots of equation (2)