Advanced higher non homogeneous 2nd order differential equation

19

In the 2019 specimen paper 2 q12 the 2nd order ode is equal to 6e^2x
Why is the trial particular solution Cx^2e^2x not just Ce^2x like in other similar examples?

Reply 1

mqb2766

21

Original post

by CossLne

In the 2019 specimen paper 2 q12 the 2nd order ode is equal to 6e^2x
Why is the trial particular solution Cx^2e^2x not just Ce^2x like in other similar examples?

Can you post a pic/link to the paper/mark scheme pls.

At a guess, the cf is something like
Ae^(2x) + Bxe^(2x)
So the pi cant be either of the two terms and must be one power greater, so
C x^2 e^(2x)

Reply 2

CossLne

OP

19

Original post

by mqb2766

Can you post a pic/link to the paper/mark scheme pls.
At a guess, the cf is something like
Ae^(2x) + Bxe^(2x)
So the pi cant be either of the two terms and must be one power greater, so
C x^2 e^(2x)

https://drive.google.com/file/d/1Yd2LL2xJnsa5SkPTFJ9X02FsG5dIRqwQ/view?usp=drivesdk
https://drive.google.com/file/d/1xhntJU1P5l9A1wHclZHtMFuWSmPDxetN/view?usp=drivesdk

Reply 3

mqb2766

21

Original post

by CossLne

https://drive.google.com/file/d/1Yd2LL2xJnsa5SkPTFJ9X02FsG5dIRqwQ/view?usp=drivesdk
https://drive.google.com/file/d/1xhntJU1P5l9A1wHclZHtMFuWSmPDxetN/view?usp=drivesdk

So what are you unsure about? Have you tried using Ce^(2x) as the pi and what did you notice when you used C x^2 e^(2x) as per the mark scheme?

Reply 4

CossLne

OP

19

Original post

by mqb2766

So what are you unsure about? Have you tried using Ce^(2x) as the pi and what did you notice when you used C x^2 e^(2x) as per the mark scheme?

Yes, using just ce^2x doesn't get the same answer... Why is the x^2 multiplying e^2x for differentiating and differentiating again? My notes just use a constant times e^2x and some equations only use a constant. What decides the fact that there's a quadratic multiplying the e^2x?

ETA: the final y equation has cx^2e^2x so there has to be a p(x)=cx^2e^2x in the trial solution. But usually the usual solution doesn't include x^2... I've not come across that and don't know what signifies to use it

Reply 5

mqb2766

21

Original post

by CossLne

Yes, using just ce^x doesn't get the same answer... Why is the x^2 multiplying e^2x for differentiating and differentiating again? My notes just use a constant times e^2x and some equations only use a constant. What decides the fact that there's a quadratic multiplying the e^2x?

If youve worked it through using Ce^(2x) you should get a 0 = 1 type equation for the PI and you should have noticed its part of the CF? Similarly if the PI was C x e^(2x). Theyre solutions to the homogeneous ode.

For the PI you want the non homogeneous ode to balance, so the left hand side must give something like Ce^(2x) after its been differentiated (twice). This will only happen if its roughly x^2 e^(2x).

Reply 6

CossLne

OP

19

Ok. So me and another had made the same error so didn't get the 0=6... Is there a quick way to determine that the extra term is needed though?

Reply 7

mqb2766

21

Original post

by CossLne

Ok. So me and another had made the same error so didn't get the 0=6... Is there a quick way to determine that the extra term is needed though?

Have you actually verified that now? You didnt post any working and it sounded like you were just reading the mark scheme. But as above, the cf is the homogeneous solution so the pi cant include those terms. It looks like youve been doing this stuff for 10 years or so, so Im assuming you know what these terms mean/represent? The x^2 multiplier is necessary to give a non zero lhs of the 2nd order ode which you should have verified and it should be reasonably clear about why that is?

Reply 8

CossLne

OP

19

Original post

by mqb2766

Have you actually verified that now? You didnt post any working and it sounded like you were just reading the mark scheme. But as above, the cf is the homogeneous solution so the pi cant include those terms. It looks like youve been doing this stuff for 10 years or so, so Im assuming you know what these terms mean/represent? The x^2 multiplier is necessary to give a non zero lhs of the 2nd order ode which you should have verified and it should be reasonably clear about why that is?

Make that 20 years! Just a bit rusty and obviously some things weren't as clearly explained as they could have been at the time. Tbh, I'm quite insulted you think I've just read the marking scheme.. I'm asking questions because I've read it with notes ! The x² multiplier stops the LHS being 0, but you only find that out from trying is my point. Is there a way to predict the x² is needed before trying ce^2x which gets the zero and then xe^2x which also gets zero (I have verified all of this but tech crashing means uploading it is not simple)

Reply 9

mqb2766

21

Original post

by CossLne

Make that 20 years! Just a bit rusty and obviously some things weren't as clearly explained as they could have been at the time. Tbh, I'm quite insulted you think I've just read the marking scheme.. I'm asking questions because I've read it with notes ! The x² multiplier stops the LHS being 0, but you only find that out from trying is my point. Is there a way to predict the x² is needed before trying ce^2x which gets the zero and then xe^2x which also gets zero (I have verified all of this but tech crashing means uploading it is not simple)

It should be in your notes and is a standard thing for solving linear odes, so you must have come across it before.

The complementary function is made of terms which solve the homogeneous ode so by definition (for this ode), both of the terms in the cf, e^(2x) and xe^(2x), satisfy
y'' - 4y' + 4y = 0
So neither of those two terms can determine the pi for the non homogeneous ode with a right hand side which is something like e^(2x), as youd get roughly 0 = e^(2x) which is obviously nonsense.

So you want to get the lhs to be roughly e^(2x) for the nonhomogeneous ode to balance and so the pi must be x^2 e^(2x) as
y = x^2 e^(2x)
y' = 2x e^(2x) + 2x^2 e^(2x)
y'' = 2 e^(2x) + 8x e^(2x) + 4x^2 e^(2x)
and subbing those into the lhs of the ode gives 2 e^(2x) and the constant multiplier associated with the pi is determined by the 2 and the rhs. The x^2 e^(2x) and the x e^(2x) terms go to zero, similar to the homogeneous solutions, which is what you want.

If the rhs e^(2x) is part of the cf, you have this scenario and the usual pi e^(2x) must be multiplied by an x or x^2 to end up with a non-zero lhs. In this case, as x e^(2x) is also part of the cf/homogeneous solution, the pi is x^2 e^(2x) which is not part of the cf/homogeneous solution.

From a very quick google about a level odes (no real knowledge of highers)
http://www.mrbartonmaths.com/resources/a%20level%20revision/2016%20Notes%20from%20Simon%20Baxter/Further%20Maths%202%20Revision%20Notes%20%20November%202014.pdf
bottom of page 18 and example 2 page 19 has this scenario. So your notes should have something similar and you should have come across it in 20 years or so. A famous example of this type of ode is where a resonant system is forced at the resonant frequency so the solution grows without bound because of the x multiplier in the pi, so something like pushing a kid on a swing or ...

(edited 8 months ago)

Advanced higher non homogeneous 2nd order differential equation

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