2nd Order, homogeneous Differential Equation

[mudweez0009]

Homework Statement

Solve d²θ/dη² + 2η(dθ/dη) = 0, to obtain θ as a function of η,
where θ=(T-T₀)/(T_s-T₀)

EDIT: I should add that this is a multi-part problem, and η is given as η=Cxt^m. We had to use that to derive the equation in question above.. So I don't 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.

Homework Equations

ay"+by'+cy=0
ar²+br+c=0
If the roots are real and different, solution is: y=ae^r₁x+be^r₂x

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!

 

[Orodruin]

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.

 

[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.

 

[Orodruin]

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.

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

 

[mudweez0009]

I don't have an equation for dθ/dη. Are you referring to the equation I have for θ in terms of T ( θ=(T-T₀)/(T_s-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.

 

[Orodruin]

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

 

[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=e^-η2

recalling that f=dθ/dη,
⇒dθ/dη = e^-η2

(This is where I may be wrong)... Rearrange and integrate again??
dθ = e^-η2dη
⇒∫dθ =∫e^-η2dη

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

 

[Orodruin]

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

 

[mudweez0009]

Okay yes, I missed the constant. So if I put that in there, the first integral would become:
f=e^-η2+C

⇒dθ/dη = e^-η2+C
⇒dθ = [e^-η2+C]dη

Integrate:
∫dθ =∫[e^-η2+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 doesn't make solving the temperature profile any easier, if that were the objective in a real-life application..

 

[Orodruin]

Okay yes, I missed the constant. So if I put that in there, the first integral would become:
f=e-η2+C

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