2nd Order, homogeneous Differential Equation

1. Apr 20, 2015

mudweez0009

1. The problem statement, all variables and given/known data
Solve d2θ/dη2 + 2η(dθ/dη) = 0, to obtain θ as a function of η,
where θ=(T-T0)/(Ts-T0)

EDIT: I should add that this is a multi-part problem, and η is given as η=Cxtm. We had to use that to derive the equation in question above.. So I dont know if this is supposed to be solved as a non-constant coefficient method or not... My method below solved it assuming η was a constant.. I can supply the whole problem as an attachment if necessary.

2. Relevant equations
ay"+by'+cy=0
ar2+br+c=0
If the roots are real and different, solution is: y=aer1x+ber2x

3. The attempt at a solution
I would assume this can just be:
θ"+2ηθ'=0
which turns to:
r2+2ηr=0

But when using the quadratic equation to get roots, I get r1=-2η, and r2=0

Plug this into the solution form and get θ=ae-2ηx

Not sure if this is right. Can someone confirm, or tell me what I'm doing wrong? Thanks!

Last edited: Apr 20, 2015
2. Apr 20, 2015

Orodruin

Staff Emeritus
If your function is a function of $\eta$, you clearly cannot make the assumption that $\eta$ is a constant.

The differential equation you have for $f(\eta) = d\theta/d\eta$ is separable.

3. Apr 20, 2015

mudweez0009

Okay, so all I know about separable equations is that you have to get like terms on the same side of the equation.
If I rearranged the original equation, it would give me:
d2θ/dθ = -2η(d2η/dη)

But I have no clue how to solve that.

4. Apr 20, 2015

Orodruin

Staff Emeritus
No, the equation for $f = d\theta/d\eta$ is separable. Solve that first and then integrate it to find $\theta$.

5. Apr 20, 2015

mudweez0009

I dont have an equation for dθ/dη. Are you referring to the equation I have for θ in terms of T ( θ=(T-T0)/(Ts-T0) )? Because I'm not sure how to differentiate that with respect to η to obtain the dθ/dη equation...
I apologize for my very rusty differential equations abilities, and I appreciate you trying to help.

6. Apr 20, 2015

Orodruin

Staff Emeritus
Make the substitution $f = d\theta/d\eta$. This gives you a differenial equation for $f$ that you have to solve.

7. Apr 20, 2015

mudweez0009

Okay I think I made progress. (crossing fingers)..
Using the substitution f=dθ/dη, that means df/dη=d2θ/dη2.
Plugging these into the equation d2θ/dη2 + 2η(dθ/dη) = 0 yields:
⇒df/dη + 2ηf =0
⇒df/dη = -2ηf
⇒(1/f)df = -2η(dη)

Integrate both sides:
⇒∫(1/f)df = ∫-2η(dη)
⇒ln(f) = -η2

Take base 'e' of both sides to get:
f=e2

recalling that f=dθ/dη,
⇒dθ/dη = e2

(This is where I may be wrong)... Rearrange and integrate again??
dθ = e2
⇒∫dθ =∫e2

θ = ½*√π*erf(η)+C (I used Wolfram Integral Calculator for this)

8. Apr 20, 2015

Orodruin

Staff Emeritus
Yes, apart from that you missed an integration constant in the first integration.

9. Apr 20, 2015

mudweez0009

Okay yes, I missed the constant. So if I put that in there, the first integral would become:
f=e2+C

⇒dθ/dη = e2+C
⇒dθ = [e2+C]dη

Integrate:
∫dθ =∫[e2+C]dη
⇒θ = ½*√π*erf(η)+η+C

I think my answer would change to include an additional η, but does that error function even make sense? I feel like that doesnt make solving the temperature profile any easier, if that were the objective in a real-life application..

10. Apr 20, 2015

Orodruin

Staff Emeritus
Not really. The constant shows up before you exponentiate the result. It is therefore a multiplicative constant rather than additive.