1st Order Linear ODE integrating factors

  • #1

Homework Statement


Hi, I submitted this question on here the other day a user suggested some topics which might help so I have went away and tried this and this is what I have came up with. I just want to know what I have so far is right also I need help with integrating the rhs of the equation.

The question is
dy/dx + 0.8y = 0.6 e-1.4x y(0)=1

I have to solve the ODE to the given condition using exact methods and evaluate the solution y for x = 0.0 to x=0.5 in steps of 0.05.


Homework Equations



dy/dx + P(x)y = Q(x)



The Attempt at a Solution



so I my working so far is

dy/dx + 0.8y = 0.6 e^-1.4x

P(x) = 0.8

IF = e ^ ∫P(x) dx
IF = e ^∫0.8 dx
= e ^ 0.8x

e0.8x* dy/dx + e0.8xy = e0.8x*0.6e-1.4x

Therefore e0.8x = ∫ e0.8x*0.6e-1.4x

This is where I get stuck. What I have done so far is this correct and what should I do to progress from this point? I don't understand how to do the integration on the rhs.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Hi, I submitted this question on here the other day a user suggested some topics which might help so I have went away and tried this and this is what I have came up with. I just want to know what I have so far is right also I need help with integrating the rhs of the equation.

The question is
dy/dx + 0.8y = 0.6 e-1.4x y(0)=1

I have to solve the ODE to the given condition using exact methods and evaluate the solution y for x = 0.0 to x=0.5 in steps of 0.05.


Homework Equations



dy/dx + P(x)y = Q(x)



The Attempt at a Solution



so I my working so far is

dy/dx + 0.8y = 0.6 e^-1.4x

P(x) = 0.8

IF = e ^ ∫P(x) dx
IF = e ^∫0.8 dx
= e ^ 0.8x

e0.8x* dy/dx + e0.8xy = e0.8x*0.6e-1.4x

Therefore e0.8x = ∫ e0.8x*0.6e-1.4x

.
What happened to the y in that last step? Also you can use rules of exponents to simplify the integrand.
 
  • #3
This is what I had
e0.8x = ∫ e0.8x*0.6e-1.4x

So should it be this instead where the multiplying exponents just add together?

e0.8xy = ∫0.6e-2.2x

What would happen now?
Would the rhs integrated become 0.6 ln(-2.2x) ?
 
  • #4
LCKurtz
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This is what I had
e0.8x = ∫ e0.8x*0.6e-1.4x

So should it be this instead where the multiplying exponents just add together?

e0.8xy = ∫0.6e-2.2x
Yes, but did you add them together correctly?
What would happen now?
Would the rhs integrated become 0.6 ln(-2.2x) ?
No. If you are taking differential equations, you should by now know your antiderivatives. Look up the antiderivative of eax in your calculus book.
 
  • #5
Ok I see that one of the exponents was -1.4 so adding that 0.8 will give us 0.6.
I have checked out my notes I just have a basic list of integrals and can't seem to find anything similar online for eax so would it be similar to eax+b where I would just miss out the plus b part? So it would just 1/0.6 e 0.6 + C? Also what would happen to the original 0.6 before the e on the rhs of the equation.
 
  • #6
LCKurtz
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Ok I see that one of the exponents was -1.4 so adding that 0.8 will give us 0.6.
I have checked out my notes I just have a basic list of integrals and can't seem to find anything similar online for eax so would it be similar to eax+b where I would just miss out the plus b part? So it would just 1/0.6 e 0.6 + C? Also what would happen to the original 0.6 before the e on the rhs of the equation.
I'm curious what level of mathematics you are taking. Are you reading ahead of the courses you are taking? You still haven't successfully added -1.4 and 0.8. And you simplify whatever you get after you solve for y.
 
  • #7
My level of maths is not too good I have to admit. I was on an easier degree course last year which had a background in the built environment it still had some maths and physics elements but wasn't as heavy on integration and differentiation. The course transferred over to a harder engineering degree and I'm totally lost on this maths module, but i'm trying to pick my way through it just to get a basic pass on the module as its the only module on my course that I have difficulty with.

So thats -0.6 ?

If not I give up ha
but thanks for your help anyway!
 
  • #8
LCKurtz
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Yes. -1.4+.8 = -0.6. I would guess that you are headed for a lot of frustration and possible failure if you are taking an engineering degree without adequate mathematics training. Good luck with that.
 
  • #9
You're right i'm pulling my hair out with it already, i'll get there in the end though! Just one last thing though. How would the anti derivative e^ax work? I'll be harassing my lecturers tomorrow haha.
 
  • #10
LCKurtz
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The derivative of eax= aeax. So the antiderivative of eax is (1/a)eax+C, assuming of course, that a isn't 0.
 
  • #11
Thanks for your time. I appreciate it.
 

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