Solving second-order ODE using an integrating factor

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.

Solutions Collecting From Web of "Solving second-order ODE using an integrating factor"

$$u”(\eta)+\frac{u'(\eta)\eta}{2\nu}=0\Longleftrightarrow$$


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


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