# 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.