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# P4 differential equation watch

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1. When tackling complementary functions with particualr integral,
in what form are the general solutions if the trial solution is form of:
• px^2+qx+r
• px+r
• psinx+qcosx
• Ae^(qx)
• Axe^(qx)
2. (Original post by keisiuho)
When tackling complementary functions with particualr integral,
in what form are the general solutions if the trial solution is form of:
• px^2+qx+r
• px+r
• psinx+qcosx
• Ae^(qx)
• Axe^(qx)
It depends on the complementory function...
it could be Ae^m1x + Be^m2x + P.I. if there are two real roots etc.

not sure if i understand what you mean....
3. Let me upload the question.

Why the general solution has so many terms?
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4. (Original post by keisiuho)
Let me upload the question.

Why the general solution has so many terms?
comp function will be the normal one (ie real distinct roots) i think.

the particular integral will be y=Ae^(3x)

if the 10e^3x was 10e^2x then you would have to use y=Axe^2x
5. also

when youre finding the complimentary function how do you know which solution to put with what letter.

for example, in that example you got +- 2 a the solutions for m.

so you can have Ae^2x + Be^-2x OR Ae^-2x + Be^2x

i suspect these things dont matter though right?

well, wrong.

working to get a particular solution the values of A and B i got were switched around with the values of the roots.

does this matter though!?
6. (Original post by kikzen)
also

when youre finding the complimentary function how do you know which solution to put with what letter.

for example, in that example you got +- 2 a the solutions for m.

so you can have Ae^2x + Be^-2x OR Ae^-2x + Be^2x

i suspect these things dont matter though right?

well, wrong.

working to get a particular solution the values of A and B i got were switched around with the values of the roots.

does this matter though!?
It doesn't matter.
7. (Original post by kikzen)
also

when youre finding the complimentary function how do you know which solution to put with what letter.

for example, in that example you got +- 2 a the solutions for m.

so you can have Ae^2x + Be^-2x OR Ae^-2x + Be^2x
isnt it only applied when there is no particular integral but only complementary function?

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