Originally Posted by
MuhTheKuh I have a problem understanding a certain step using the Variation of Parameters Method.
take for example
y''+y=sec(x)tan(x)
Auxiliary Equation: mē+1=0 thus m= +/- 1
CF: Y1= Acos(x)+Bsin(x)
then I let y=A(x)cos(x)+B(x)sin(x)
and differentiate to
y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later)
differentiating again
y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x)
Substituting the yields now:
-A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step)
so now I have to solve for A' and/or B'
I have no idea how to do this and when I checked the answer booklet it explained the next step by:
* cos(x) - ** sin(x): A'=tanē(x) = 1 - sec(x) A= x - tan(x) + C
and
* sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D
I fully understand how to get up to the point were it says
* cos(x) - ** sin(x)
and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked (*) differentiations to A'= or B'=
I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet.
Thank you guys for your time, I appreciate it.