FP3 Diff Equations Section Watch

tom29whu
Badges: 2
Rep:
?
#1
Report Thread starter 7 years ago
#1
Can someone quickly summarise the definitions different names for things in the differential equations section of FP3, what form they are generally in, and very briefly, how you work them out

Such as;
Particular Integral
General Solution
Integrating Factor
Complementary Function

And any others if I've missed them.

I would really appreciate it if someone could summarise this for me, I would be EXTREMELY grateful.

Thanks,
Tom
0
reply
mr tim
  • Forum Helper
Badges: 17
Rep:
?
#2
Report 7 years ago
#2
(Original post by tom29whu)
Can someone quickly summarise the definitions different names for things in the differential equations section of FP3, what form they are generally in, and very briefly, how you work them out

Such as;
Particular Integral
General Solution
Integrating Factor
Complementary Function

And any others if I've missed them.

I would really appreciate it if someone could summarise this for me, I would be EXTREMELY grateful.

Thanks,
Tom
Particular Intergral - you look at the RHS of the original equation [the f(x) part], and you decide what to make y_p equal to. This is what you do:

if f(x) is a polynomial with degree n - P.I. is a polynomial with degree n
Eg: x^2 would become Px^2+Qx^+R

if f(x) is e^{kx} - PI is Pe^kx
eg: 6e^{7x} is Pe^{7x}

if f(x) is Sin(kx) and/or Cos(kx) - PI is Psin(kx) + Qcos(kx)

Then you would differentiate twice the P.I. to get \frac{dy}{dx} and \frac{d^2y}{dx^2} and sub it in the original equation.

Complementary function - you put A\frac{d^2y}{dx^2}+B\frac{dy}{dx  }+Cy=0

General Solution - if the discriminant of the auxiliary equation of the C.F. is not equal to 0, then the general solution is in the form Ae^Px, and Be^Qx where P and Q are the factors of the auxiliary equation. If the discriminate of the auxiliary eqn. of the C.F. is equal to 0, the Gen solution will be in the form (Ax+B)e^(kx), where k is the factor of that Aux equation. When the RHS is f(x) we do y= C.F+P.I

Integrating Factor if the equation is in the form \frac{dy}{dx}+P(x)y=Q(x) then we use the I.F. method. [you may need to rearrange to get it in this form if you can]. We basically intergrate the p(x) part and then work out e^(the intergral of p(x). Then we multiply that result by the original equation.

Hope you understand these.
1
reply
Farhan.Hanif93
Badges: 18
Rep:
?
#3
Report 7 years ago
#3
(Original post by mr tim)
Particular Intergral - you look at the RHS of the original equation [the f(x) part], and you decide what to make y_p equal to. This is what you do:

if f(x) is a polynomial with degree n - P.I. is a polynomial with degree n
Eg: x^2 would become Px^2+Qx^+R

if f(x) is e^{kx} - PI is Pe^kx
eg: 6e^{7x} is Pe^{7x}

if f(x) is Sin(kx) and/or Cos(kx) - PI is Psin(kx) + Qcos(kx)

Then you would differentiate twice the P.I. to get \frac{dy}{dx} and \frac{d^2y}{dx^2} and sub it in the original equation.

Complementary function - you put A\frac{d^2y}{dx^2}+B\frac{dy}{dx  }+Cy=0

General Solution - if the discriminant of the auxiliary equation of the C.F. is not equal to 0, then the general solution is in the form Ae^Px, and Be^Qx where P and Q are the factors of the auxiliary equation. where the RHS is f(x) we do y= C.F+P.I

Integrating Factor if the equation is in the form \frac{dy}{dx}+P(x)y=Q(x) then we use the I.F. method. [you may need to rearrange to get it in this form if you can]. We basically intergrate the p(x) part and then work out e^(the intergral of p(x). Then we multiply that result by the original equation.

Hope you understand these.
You've missed quite a few vital parts out. For example, what happens to your choice of your trial function for the particular integral when the complementary function is of the form of one of the trial functions you have posted?
0
reply
mr tim
  • Forum Helper
Badges: 17
Rep:
?
#4
Report 7 years ago
#4
(Original post by Farhan.Hanif93)
You've missed quite a few vital parts out. For example, what happens to your choice of your trial function for the particular integral when the complementary function is of the form of one of the trial functions you have posted?
yeah.. Haven't really learnt the full D.E 2nd order yet...
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (100)
40.32%
No - but I will (12)
4.84%
No - I don't want to (16)
6.45%
No - I can't vote (<18, not in UK, etc) (120)
48.39%

Watched Threads

View All