Por supuesto, hay una familia de soluciones simples de dos parámetros: $y(x) = a + b x$ .
Podrías buscar soluciones en serie:
$$\eqalign{y \left( x \right) &=a+bx+c{x}^{2}-{\frac {ac}{3}}{x}^{3}+ \left( { \frac {{a}^{2}c}{12}}-{\frac {bc}{12}} \right) {x}^{4}+ \left( -{ \frac {{a}^{3}c}{60}}+{\frac {bac}{20}}-{\frac {{c}^{2}}{30}} \right) {x}^{5}\cr &+ \left( {\frac {{c}^{2}a}{36}}+{\frac {{a}^{4}c}{360}}-{\frac {b{a}^{2}c}{60}}+{\frac {{b}^{2}c}{120}} \right) {x}^{6}+ \left( -{ \frac {4\,{a}^{2}{c}^{2}}{315}}+{\frac {11\,b{c}^{2}}{1260}}-{\frac {{ a}^{5}c}{2520}}+{\frac {b{a}^{3}c}{252}}-{\frac {{b}^{2}ac}{168}} \right) {x}^{7}\cr &+ \left( {\frac {{a}^{3}{c}^{2}}{240}}-{\frac {ba{c}^{ 2}}{120}}+{\frac {{a}^{6}c}{20160}}-{\frac {b{a}^{4}c}{1344}}+{\frac { {b}^{2}{a}^{2}c}{448}}-{\frac {{b}^{3}c}{1344}}+{\frac {11\,{c}^{3}}{ 5040}} \right) {x}^{8}\cr &+ \left( -{\frac {13\,{c}^{3}a}{5040}}-{\frac { 11\,{a}^{4}{c}^{2}}{10080}}+{\frac {4\,b{a}^{2}{c}^{2}}{945}}-{\frac { 43\,{b}^{2}{c}^{2}}{30240}}-{\frac {{a}^{7}c}{181440}}+{\frac {b{a}^{5 }c}{8640}}-{\frac {{b}^{2}{a}^{3}c}{1728}}+{\frac {{b}^{3}ac}{1728}} \right) {x}^{9}\cr & + \ldots} $$
En particular, el caso $a=b=0$ es bastante agradable:
$$ \eqalign{y \left( x \right) &=c{x}^{2}-{\frac {{c}^{2}}{30}}{x}^{5}+{\frac {11 \,{c}^{3}}{5040}}{x}^{8}-{\frac {5\,{c}^{4}}{33264}}{x}^{11}+{\frac { 9299\,{c}^{5}}{908107200}}{x}^{14}-{\frac {1272379\,{c}^{6}}{ 1852538688000}}{x}^{17}\cr& +{\frac {19241647\,{c}^{7}}{422378820864000}}{x }^{20}-{\frac {33675995567\,{c}^{8}}{11220493376252160000}}{x}^{23}\cr &+{ \frac {17241364408921\,{c}^{9}}{87519848334766848000000}}{x}^{26}-{ \frac {1375703592341009\,{c}^{10}}{106599175271746020864000000}}{x}^{ 29}+\ldots} $$
EDIT: El coeficiente de $x^{3n-1}$ aquí está $- a(n-1) (-c)^n/(3n-1)!$ donde $a(n)$ es la secuencia OEIS A018893 .
EDIT: Véase Bluman y Anco, "Symmetry and Integration Methods for Differential Equations", sec. 3.4.2, donde la ecuación de Blasius se reduce a una ecuación de primer orden.