1. Reduction to Bessel's equation

Hi Guys, just hoping someone might be able to shed a little light on the following problem. I'm working my through Kreyzig's Advanced Engineerig Maths and have come unstuck. the given ODE is x^2y''-5xy'+9(x^6-8)y=0
(y=(x^3)*u) ((x^3)=z). The solutions manual gives the following for y prime.
[1] dy/dx=d/dx (x^3)u = 3(x^2)u+(x^3)du/dz*3x^2 = (3x^5)u'+(3x^2)u
Until know I thought I understood the chain rule, but obviously I don't have a clue. Any ideas on how he went from d/dx (x^3)u to (3x^5)u'+(3x^2)u would be very much appreciated.
Cheers DynamicDan

2. Originally Posted by DynamicDan
Hi Guys, just hoping someone might be able to shed a little light on the following problem. I'm working my through Kreyzig's Advanced Engineerig Maths and have come unstuck. the given ODE is x^2y''-5xy'+9(x^6-8)y=0
(y=(x^3)*u) ((x^3)=z). The solutions manual gives the following for y prime.
[1] dy/dx=d/dx (x^3)u = 3(x^2)u+(x^3)du/dz*3x^2 = (3x^5)u'+(3x^2)u
Until know I thought I understood the chain rule, but obviously I don't have a clue. Any ideas on how he went from d/dx (x^3)u to (3x^5)u'+(3x^2)u would be very much appreciated.
Cheers DynamicDan
I'm sorry but I can't follow what you've posted. What does (y=(x^3)*u) ((x^3)=z) mean?

Do you mean that you first make the substitution $\displaystyle y=x^3 u \Rightarrow \frac{dy}{dx} = 3x^2 u + x^3 \frac{du}{dx}$ ?

And then make the substitution $\displaystyle z = x^3 \Rightarrow \frac{du}{dx} = \frac{du}{dz} \cdot \frac{dz}{dx} = \frac{du}{dz} 3x^2$ ?

In which case $\displaystyle \frac{dy}{dx} = 3x^2 u + x^3 \frac{du}{dx} = 3x^2 u + x^3 \frac{du}{dz} 3x^2 = 3x^2 u + 3x^5 \frac{du}{dz}$ ?

In which case the term $\displaystyle 5 x \frac{dy}{dx}$ in the given DE becomes $\displaystyle 15x^3 u + 15 x^6 \frac{du}{dz} = 15 z u + 15 z^2 \frac{du}{dz}$ ......

3. Hi Mr Fantastic, sorry I wasn't clearer, but you got it anyway. Thanks. It seems clear now you stepped it out for me.
Regards Dan

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ODE engineerig

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