1st Order Linear ODE integrating factors

[andycampbell1]

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

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

Therefore e^0.8x = ∫ e^0.8x*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.

 

[LCKurtz]

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

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

Therefore e^0.8x = ∫ e^0.8x*0.6e^-1.4x

.

What happened to the y in that last step? Also you can use rules of exponents to simplify the integrand.

 

[andycampbell1]

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?

e^0.8xy = ∫0.6e^-2.2x

What would happen now?
Would the rhs integrated become 0.6 ln(-2.2x) ?

 

[LCKurtz]

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?

e^0.8xy = ∫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 e^ax in your calculus book.

 

[andycampbell1]

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 e^ax so would it be similar to e^ax+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.

 

[LCKurtz]

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 e^ax so would it be similar to e^ax+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.

 

[andycampbell1]

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 that's -0.6 ?

If not I give up ha
but thanks for your help anyway!

 

[LCKurtz]

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.

 

[andycampbell1]

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.

 

[LCKurtz]

The derivative of e^ax= ae^ax. So the antiderivative of e^ax is (1/a)e^ax+C, assuming of course, that a isn't 0.

 

[andycampbell1]

Thanks for your time. I appreciate it.