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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

I will get a pen and paper

- anonymous

thank you I know its alot of work

- anonymous

ok I solved it

- anonymous

so do you know the technique of integrating factor?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

no problem, what is the integrating factor here?

- anonymous

ok

- anonymous

after we integrate x dx on the other side

- anonymous

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

- anonymous

that is -2 here

- anonymous

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

- anonymous

-2x

- anonymous

Im sorry the computer wont let me type

- anonymous

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

- anonymous

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

- anonymous

do you recognise the full derivative now?

- anonymous

yes it is coming back to me

- anonymous

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

- anonymous

this is harder but my favourite :)

- anonymous

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

- anonymous

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,\[a \lambda ^{2}+b \lambda +c=0\]

- anonymous

where lambda=e^x

- anonymous

or in the form of e^x

- anonymous

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

- anonymous

do you know how to?

- anonymous

I will be back in 5 min

- anonymous

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

- anonymous

you can only solve this with complex numbers

- anonymous

thank ufor your help

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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..

- anonymous

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

- anonymous

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

- anonymous

what is the end bit? that isnt clear

- anonymous

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

- anonymous

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

- anonymous

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

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