2nd order diff equation Q

For b), how would i know what form of PI to use?

use as PI

whatever they suggest if they say so

or

u=Psin2x+Qcos2x (you will get Q=0, which is ok/you should not lose marks)

or

with practice and experience you will be able to see in an ODE like this one, that the cos2x term is not needed

WOuld a PI of y = axcos2x work?
Its says in my book that if f(x) = CsinEx
And CF is of form Acos(Ex) + Bsin(Ex)

Then use PI y = axcos(Ex)

Thanks, i found the some relevant questions from the website you introduced me too.
Was of great help, thanks alot!

WOuld a PI of y = axcos2x work?
Its says in my book that if f(x) = CsinEx
And CF is of form Acos(Ex) + Bsin(Ex)

Then use PI y = axcos(Ex)

You're getting confused about 2 different things here (as I think you were in your thread the other day).

The choice of PI is normally fairly obvious for A level.

E.g. if the RHS is an exponential, try an exponential for the PI.
if the RHS is a polynomial of degree 5, try a polynomial of degree 5 for the PI
if the RHS contains cos(ax) or sin(ax), try a combination of Acos(ax) + Bsin(ax) for the PI.

The reason you need a combination in the third case above is because if y contains sin (or cos) then the derivative y' will contain cos (or sin) and the second derivative will contain sin (or cos), so if your equation contains y'', y' and y terms then you need both sin and cos in the PI otherwise you'll get terms that don't cancel out when you differentiate and add together.

In the example you've posted here, you might think you need both terms but note that there is no y' on the LHS. Therefore you can get away with Asin2x as the PI because y'' also looks like Bsin2x and you never get a cos2x term introduced on the LHS!

The point you found about trying xsin2x as a PI is irrelevant for this equation! You only need to worry about multiplying by x if the "obvious" PI would match up with a term in the CF. E.g. if the CF contains a sin2x, it's no good trying sin2x in the PI because it will just give 0 when you put it in the original DE! In your case, you have cos x and sin x in the CF, so a PI of Asin2x is fine by itself.

Ok but i read it from the book.

I'm not sure what you're taking issue with The book agrees with what I've said.

Your CF involves sin x and cos x terms. Your DE has an f(x) which involves sin2x which isn't the same as the CF. Therefore the PI is just ksin2x. You would only need kxsin2x if the CF involved sin2x.

yeah i just posted again, sorry. I understand now, thanks for the help!

I'm not sure what you're taking issue with The book agrees with what I've said.

Your CF involves sin x and cos x terms. Your DE has an f(x) which involves sin2x which isn't the same as the CF. Therefore the PI is just ksin2x. You would only need kxsin2x if the CF involved sin2x.

So its basically just a case of whether f(x) is contained in the CF or not.
If it is then multiply by x and use that PI. If not use a constant?

ThankS!

Basically yes!

It's worth trying to see for yourself what goes wrong if you don't do that. E.g. if f(x) = sin x in your equation, what happens when you try a PI of ksin x?