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'))

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)