Solve the ODE $$ \frac{\partial^{2} u }{\partial \eta^{2}} + \frac{\eta}{2\nu} \frac{\partial}{\partial\eta} = 0 $$

The book uses integrating factor = $ e^{\int\frac{\eta}{2\nu} d\eta}$

Can someone explain from here, how to proceed?

I get $ IF = e^\frac{\eta^{2}}{4\nu}$. Multiply this with the ODE. Not sure what to do next? Because the answer the book gives is $$ \frac{\partial u}{\partial \eta} = A e^{\frac{-\eta^{2}}{4/nu}}$$ and I don't understand how to get this.

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    $\begingroup$ The idea of IF is that after multiplication you will get something of the form $fg'+f'g = (fg)'$. $\endgroup$ – user147263 Jan 24 '16 at 1:40
  • $\begingroup$ \begin{align} u'' + \frac{n}{2\nu}u' &= 0 \\ \implies \bigg[ u' \exp \bigg( \frac{n^{2}}{4\nu} \bigg) \bigg]' &= 0 \\ \implies u' \exp \bigg( \frac{n^{2}}{4\nu} \bigg) &= A \\ \implies u' &= A \exp \bigg( \frac{-n^{2}}{4\nu} \bigg) \end{align} $\endgroup$ – Mattos Jan 24 '16 at 3:57


Let $u'(\eta)=w(\eta)$, which gives $u''(\eta)=w'(\eta)$:

$$w'(\eta)=-\frac{\eta w(\eta)}{2\nu}\Longleftrightarrow$$ $$\frac{2vw'(\eta)}{w(\eta)}=-\eta\Longleftrightarrow$$ $$\int\frac{2\nu w'(\eta)}{w(\eta)}\space\text{d}\eta=\int-\eta\space\text{d}\eta\Longleftrightarrow$$ $$2\nu\ln(w(\eta))=-\frac{\eta^2}{2}+\text{C}\Longleftrightarrow$$ $$w(\eta)=e^{-\frac{\eta^2-2\text{C}}{4\nu}}\Longleftrightarrow$$ $$w(\eta)=e^{-\frac{\eta^2+\text{C}}{4\nu}}\Longleftrightarrow$$ $$u'(\eta)=e^{-\frac{\eta^2+\text{C}}{4\nu}}\Longleftrightarrow$$ $$u(\eta)=\int e^{-\frac{\eta^2+\text{C}}{4\nu}}\space\text{d}\eta$$


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