I have used the steam functions $u = \psi_{y}$ and $v = -\psi_{x}$ to transform the momentum equations to the following form $$\rho\left(\psi_{y}\psi_{xy} - \psi_{x}\psi_{yy}\right)=-p_{x}+\mu\left(\psi_{xxy}+\psi_{yyy} \right) $$ $$\rho\left(-\psi_{y}\psi_{xx} + \psi_{x}\psi_{xy}\right)=-p_{y}+\mu\left(\psi_{xxx}+\psi_{xyy} \right) $$
I eliminated the pressure term via cross-differentiation and have obtained the final form of the equation to be as follows $$\rho\left(\psi_{y}\left(\psi_{xyy}+\psi_{xxx} \right) - \psi_{x}\left(\psi_{yyy}+\psi_{xxy} \right) \right) = \mu\left(\psi_{xxxx} +\psi_{yyyy} + 2\psi_{xxyy} \right)\tag{1}, $$ which can be rewritten with $\nabla = \left<\partial_{x},\partial_{y} \right>$ as follows $$\rho\left(\psi_{y}\nabla^{2}\psi_{x} - \psi_{x}\nabla^{2}\psi_{y} \right) = \mu\nabla^{4}\psi$$
Now, this is a leading edge problem of a fluid flowing over a flat plate with viscous forces dominating. The self-similar variable was found to be $$\eta = \frac{y}{x}\tag{2},$$ and the self-similar stream function has the following form $$f(\eta) = \frac{\psi}{Ux}.\tag{3} $$
Substituting $(2)$ and $(3)$ into $(1)$ transforms the PDE to an ODE a follows $$\left(1+\eta^{2}\right)^{2}f_{\eta\eta\eta\eta}+8\eta\left(1+\eta^{2}\right)f_{\eta\eta\eta} + 4\left(1+3\eta^{2}\right)f_{\eta\eta} + Re\left[2\eta ff_{\eta}+\left(1+\eta^{2}\right)\left(ff_{\eta\eta\eta} + f_{\eta}f_{\eta\eta}\right)\right]=0,\tag{4}$$ where $Re=\frac{\rho Ux}{\mu}$
Could someone please show me how this transformation is done. I tried proceeding via the chain rule and was able to compute all the derivatives but failed to obtain the final form as shown in $(4)$. Also, could anyone please derive the final ode if $f(\eta,Re)=\frac{\psi}{Ux}$? Any help is appreciated.
