Derivation of Differential Equation with Parameter Pr and Solution Integration

[Saladsamurai]

Homework Statement

I am going through a derivation in my text and we come across the following diff EQ:

θ'' / θ' = Pr* f''' / f''​

where Pr = constant (it ia a parameter) and both θ and f are functions of the same independent variable η.

The Attempt at a Solution

The result we arrive at is

$$\theta(\eta,Pr) =C_1 \int_0^\eta [f'']^{Pr}\,d\eta +C_2\qquad(2)$$​

and he the author says we get to that results by "integrating (1) twice ..."

I guess I am a little confused as to how to integrate this? Do I need to do it by parts? Or move the dη's around or what? Any thoughts? Am I making this too difficult?

 

[dextercioby]

No, it's just a trick to observe that the initial differential relation

$$\frac{\theta''}{\theta'} = P_{r} \frac{f'''}{f''}$$

is nothing but

$$(\ln \theta')' = P_{r} (\ln f'')'$$

in disguise.

Integrating once we get

$$\ln \theta ' = P_{r} \ln f'' + \ln C$$

from which

$$\theta' = C (f'')^{P_{r}}$$

Integrate now from 0 to $\eta$ and you'll get the answer in the book.

 

[Saladsamurai]

No, it's just a trick to observe that the initial differential relation

$$\frac{\theta''}{\theta'} = P_{r} \frac{f'''}{f''}$$

is nothing but

$$(\ln \theta')' = P_{r} (\ln f'')'$$

in disguise.

Integrating once we get

$$\ln \theta ' = P_{r} \ln f'' + \ln C$$

from which

$$\theta' = C (f'')^{P_{r}}$$

Integrate now from 0 to $\eta$ and you'll get the answer in the book.

Oh wow! Nice catch Thanks dextercioby