franky2727 said:
ok so I am looking at innitial value problems such as y''-3y'+2y=cosx for instance for this my Yp needs to be acosx +bsinx then yp' and yp'' etc
my question is what are the yp's of the following right hand sides
=ex yp=axex??
=ex+2 yp=axex??
thanks
Why would you ask? You can differentiate can't you? If y= axe
x, then y'= ae
x+ ax
x and y"= 2ae
x+ axe
x.
Putting those into the equation, 2ae
x+ axe
x- 3(ae
x+ ax
x)+ 2(axe
x)= (1- 3+ 2)axe
x+ (2-3)ae
x= -ae
x. Choosing a= -1 will give you e
x but certainly NOT e
x+ 2.
You cannot decide what functions to try without looking at the solutions to the homogenous equation, y"- 3y'+ 2y= 0, which has characteristic equation r
2- 3r+ 2= (r- 1)(r- 2). that tells you that the general solution to the homogenous equation is Ce
x+ De
2x and I presume that is why you chose a
xe
x for your yp.
However, any constant, like 2, can be written as 2e
0x since e
0= 1. Since r= 0 is not a solution to the characteristic equation you should be looking for a solution of the form be
0x= b. In other words, as long as a constant is not already a solution to the homogeneous equation, you should try 'b' as a solution.
You should try yp= axe
x+ b.
By the way, one of the crucial points about "linear" equations is that you can solve the individual parts and then combine them. That's why you can solve the homogeneous equation to get the general solution and then combine with specific solutions to the entire equation. It also means that you can treat different parts of the right hand side separately:
To find a particular solution to y"- 3y'+ 2y= e
x you would try yp= axe
x and to find a particular solution to y"- 3y'+ 2y= 2 you would try yp= b, a constant. To find a particular solution to y"- 3y'+ 2y= e
x+ 2, you add those two particular solutions.
