Question

anyone a pro with differential equations? It has been too long and I need some help

Answer 1

I feel ok with them, not pro but hopefully I can help

Answer 2

\[y'''+y=0 \] where y(o)=0 , y'(0) =1, y''(0)= 0 thank you please help

Answer 3

sorry I cannot help, havent done 3rd order ones.

Answer 4

Ok how about find all solutions to y'-2y=1 thank you for trying

Answer 5

you can multiply with an integrating factor, do you remember that?

Answer 6

I will get a pen and paper

Answer 7

thank you I know its alot of work

Answer 8

ok I solved it

Answer 9

so do you know the technique of integrating factor?

Answer 10

you r the best lets see if I get it :)

Answer 11

you want to make a full derivative out of the left hand side

Answer 12

for this you need to have something like dy/dx+ some y=whatever

Answer 13

my computer is slowing down sorry ...yes I remeber that

Answer 14

no problem, what is the integrating factor here?

Answer 15

ok

Answer 16

after we integrate x dx on the other side

Answer 17

it is \[e ^{\int\limits_{}^{} the multiplier of y}dx\]

Answer 18

that is -2 here

Answer 19

the integral of -2 is -2x (you dont need the constant here)
so multiply through by \[e ^{-2}\]

Answer 20

-2x

Answer 21

Im sorry the computer wont let me type

Answer 22

It keeps stopping but I am here reading...Yes I remeber now the -2x

Answer 23

the result is
\[e ^{-2x}\] dy/dx -2 e^{-2x} y=e ^{-2x}

Answer 24

do you recognise the full derivative now?

Answer 25

yes it is coming back to me

Answer 26

what about when we have a trig function like y''+4y=cos x

Answer 27

this is harder but my favourite :)

Answer 28

better your fav than mine :) :) i have not done these in 17 years it is so hard to remember you r a big help

Answer 29

first you have to think of the homogeneous equation (meaning y''+4y=0)
you have to find the general solution for that, for this we use the auxiliary equation. that just says that ay''+by'+cy=0
and write a quadratic equation for it,\[a \lambda ^{2}+b \lambda +c=0\]

Answer 30

where lambda=e^x

Answer 31

or in the form of e^x

Answer 32

here you will have to solve y^2+4=0

Answer 33

do you know how to?

Answer 34

I will be back in 5 min

Answer 35

i have no idea Im trying to follow along with you and my book but so confused I need to watch a tutorial i think and i will be bac

Answer 36

you can only solve this with complex numbers

Answer 37

thank ufor your help

Answer 38

you had enough?
How is it that after 17 years you are doing this again?

Answer 39

http://www.analyzemath.com/calculus/Differential_Equations/second_order.html
This page explains it quite well

Answer 40

i am a teacher but needed 2 credits of advanced math i truly didnt want to take up your time

Answer 41

I will have an exam about this in 2weeks so it is not a waste of time for me

Answer 42

I need alot of explanation and my computer is not cooperating ...to be honest this work is due tomorrow and my brain is blank we can continue if u want to help me..

Answer 43

well it is up to you, I am free now for an hour

Answer 44

Consider the equation y' + (cos x) y= e^-sinx find the solution that satisfies \[\phi\](\[\pi\] ) = pi

Answer 45

what is the end bit? that isnt clear

Answer 46

I guess you have to use the integrating factor here as well. it will be e^integral(cosx)
so e^sinx.

Answer 47

that way you get \[(e ^{sinx}y)'=1\]

Answer 48

so the LHS= x+C
y=(x+C)/e^sinx