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Fp2 second order differential equations Particular Integral adding an x question

Mutleybm1996

Hi all,
With the attached question, when they find the particular integral they use

Original post by Mutleybm1996

Could someone possibly explain this please.
Thanks!

Looking at your auxilliary equation, your two roots are

\{-1+3i, -1-3i\}

, and that set does not include

-1

; it's a completely different number.

Although

e^{-x}

does appear in one form of the given complementary function, it's multiplied by sin/cos and works out to give the two values previously stated. That "-1" is flying under false colours.

Ae^{(-1+3i)x}+Be^{(-1-3i)x}

can be rewritten as:

e^{-x}(Ae^{3ix}+Be^{-3ix})

Which for suitable choices of C,D can be rewritten as:

e^{-x}(C\cos 3x+D\sin 3x)

Edit: Missed out a lot of x's - corrected.

(edited 8 years ago)

Reply 3

Zenarthra

Original post by ghostwalker

Looking at your auxilliary equation, your two roots are

\{-1+3i, -1-3i\}

, and that set does not include

-1

; it's a completely different number.

Although

e^{-x}

does appear in one form of the given complementary function, it's multiplied by sin/cos and works out to give the two values previously stated. That "-1" is flying under false colours.

Ae^{-1+3i}+Be^{-1-3i}

can be rewritten as:

e^{-1}(Ae^{3i}+Be^{-3i})

Which for suitable choices of C,D can be rewritten as:

e^{-1}(C\cos 3x+D\sin 3x)

So when should you actually multiply by x?

Reply 4

rayquaza17

Original post by Zenarthra

So when should you actually multiply by x?

When you have repeated roots for the CF.

Reply 5

Zenarthra

Original post by rayquaza17

When you have repeated roots for the CF.

Only when you have repeated roots?
What about when roots are not equal but PI is contained in the CF?

Reply 6

rayquaza17

Original post by Zenarthra

Only when you have repeated roots?
What about when roots are not equal but PI is contained in the CF?

If the CF is the same form as the RHS of the ODE, then yeah multiply it by t.
Like in example 1 on here: http://www.ncl.ac.uk/students/mathsaid/resources/academic/ode_resonance.htm

But as mentioned above, here the CF is in the form

e^{-x}sin(3x)+e^{-x}cos(3x)

which isn't the same form as

e^{-x}

Reply 7

davros

Original post by Zenarthra

Only when you have repeated roots?
What about when roots are not equal but PI is contained in the CF?

If you're trying to find the CF only, repeated roots indicates that you can't get two linearly independent solutions from the AE, so for a CF with two constants to find, you need an exponential from the repeated root plus a term that is that exponential multiplied by x to give you another independent solution.

If you're trying to find the PI, and the "obvious" solution would be a constant times an exponential, but the CF already contains that exponential multiplied by a constant, then the "obvious" PI won't work because if you plug it into the DE you'll just get 0, not the term you want on the RHS! Therefore, you try the "next simplest" PI which is a constant times x times the exponential.

However, something like

Ae^{ax} \cos{bx}

in the CF isn't a "constant times the exponential" it's constant times function times exponential, so the same problem doesn't arise! Therefore you can have