Solve Question (D^2 + a^2 )y=Sec x: Step-by-Step Guide

[mohdfasieh]

hello genius guys,
can u people tell me the method of solving this question

QUESTION:1
(D^2 + a^2 )y=Sec x
STEP 1:
y=sec x/((D+ai)(D-ai))
STEP2:
y=1/(d+ai) *\exp ^aix \int Secax \exp ^-aix dx


please please tell me any method of solving this question i have a test tomorrow please tell me the solution

 

[benorin]

Is D is a differential operator?

 

[mohdfasieh]

yes D is a differential operator

 

[HallsofIvy]

Benorin, that's a fairly standard notation.

Mohdfasieh, it looks like you are trying to put it into form for a transform solution. I don't much like those.

Here's how I would do it: The associated homogeneous d.e. is
(D²+ a^2)y= 0 which has solutions (for a not 0)
y(x)= C cos(ax)+ D sin(ax).

Now find a solution to the entire equation by assuming a solution of the form y(x)= u(x)cos(ax)+ v(x)sin(ax). Differentiating,
y'= u' cos(ax)- au sin(ax)+ v'sin(ax)+ avcos(ax).

Since there are, in fact, many such solutions, reduce the search to those for which u' cos(ax)+ v' sin(ax)= 0.

Now we have y'= -au sin(ax)+ av cos(ax) so y"= -au' sin(ax)- a² cos(ax)+ av' cos(ax)- a²v sin(ax)

Putting those into the equation, y"+ a²y= -au' sin(ax)- a²u cos(ax)+ av' cos(ax)- a²v sin(ax)+ a²u cos(ax)+ a²v sin(ax)= -au' sin(ax)+ a v' cos(ax)= sec(ax).

That, together with u' cos(ax)+ v' sin(ax)= 0 gives to equations to solve for u' and v' which then can (theoretically) be integrated.