anonymous
  • anonymous
anyone a pro with differential equations? It has been too long and I need some help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I feel ok with them, not pro but hopefully I can help
anonymous
  • anonymous
\[y'''+y=0 \] where y(o)=0 , y'(0) =1, y''(0)= 0 thank you please help
anonymous
  • anonymous
sorry I cannot help, havent done 3rd order ones.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Ok how about find all solutions to y'-2y=1 thank you for trying
anonymous
  • anonymous
you can multiply with an integrating factor, do you remember that?
anonymous
  • anonymous
I will get a pen and paper
anonymous
  • anonymous
thank you I know its alot of work
anonymous
  • anonymous
ok I solved it
anonymous
  • anonymous
so do you know the technique of integrating factor?
anonymous
  • anonymous
you r the best lets see if I get it :)
anonymous
  • anonymous
you want to make a full derivative out of the left hand side
anonymous
  • anonymous
for this you need to have something like dy/dx+ some y=whatever
anonymous
  • anonymous
my computer is slowing down sorry ...yes I remeber that
anonymous
  • anonymous
no problem, what is the integrating factor here?
anonymous
  • anonymous
ok
anonymous
  • anonymous
after we integrate x dx on the other side
anonymous
  • anonymous
it is \[e ^{\int\limits_{}^{} the multiplier of y}dx\]
anonymous
  • anonymous
that is -2 here
anonymous
  • anonymous
the integral of -2 is -2x (you dont need the constant here) so multiply through by \[e ^{-2}\]
anonymous
  • anonymous
-2x
anonymous
  • anonymous
Im sorry the computer wont let me type
anonymous
  • anonymous
It keeps stopping but I am here reading...Yes I remeber now the -2x
anonymous
  • anonymous
the result is \[e ^{-2x}\] dy/dx -2 e^{-2x} y=e ^{-2x}
anonymous
  • anonymous
do you recognise the full derivative now?
anonymous
  • anonymous
yes it is coming back to me
anonymous
  • anonymous
what about when we have a trig function like y''+4y=cos x
anonymous
  • anonymous
this is harder but my favourite :)
anonymous
  • 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
  • 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
  • anonymous
where lambda=e^x
anonymous
  • anonymous
or in the form of e^x
anonymous
  • anonymous
here you will have to solve y^2+4=0
anonymous
  • anonymous
do you know how to?
anonymous
  • anonymous
I will be back in 5 min
anonymous
  • 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
  • anonymous
you can only solve this with complex numbers
anonymous
  • anonymous
thank ufor your help
anonymous
  • anonymous
you had enough? How is it that after 17 years you are doing this again?
anonymous
  • anonymous
http://www.analyzemath.com/calculus/Differential_Equations/second_order.html This page explains it quite well
anonymous
  • anonymous
i am a teacher but needed 2 credits of advanced math i truly didnt want to take up your time
anonymous
  • anonymous
I will have an exam about this in 2weeks so it is not a waste of time for me
anonymous
  • 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
  • anonymous
well it is up to you, I am free now for an hour
anonymous
  • anonymous
Consider the equation y' + (cos x) y= e^-sinx find the solution that satisfies \[\phi\](\[\pi\] ) = pi
anonymous
  • anonymous
what is the end bit? that isnt clear
anonymous
  • anonymous
I guess you have to use the integrating factor here as well. it will be e^integral(cosx) so e^sinx.
anonymous
  • anonymous
that way you get \[(e ^{sinx}y)'=1\]
anonymous
  • anonymous
so the LHS= x+C y=(x+C)/e^sinx

Looking for something else?

Not the answer you are looking for? Search for more explanations.