Differential equation $y' +y^2 = \frac{15}{4x^2}$ Could you please assist me with the method of this question? I haven't been able to find similar examples. 
The information in the question is: Knowing that $u(x) = \frac52\cdot x^{-1}$ is the particular solution of $y' +y^2 = \frac{15}{4x^2}$ find the solution of $y=y(x)$ that satisfies $y(1) = 0$.
The answer is $$\frac{15\cdot x^4-15}{6\cdot x^5+10\cdot x}$$
 A: You are probably expected to take the given solution $u(x)$ to transform the Ricatti equation into a Bernoulli equation for $v$ in $y=u+v$,
$$
u'+v'+u^2+2u_0v+u_0^2=r=u'+u^2
\\~\\
\implies v'+2uv+v^2=0\implies -(v^{-1})'+2u(v^{-1})+1=0
$$
which then is as usual transformed into a linear DE for some power of $v$.

Another method is to set $y=\frac{u'}{u}$. This results in the linear DE
$$
4x^2u''(x)-15u(x)=0
$$
which is an Euler-Cauchy DE. This has solutions of the form $x^m$ with characteristic polynomial
$$
0=4m(m-1)-15=(2m-1)^2-16=(2m-5)(2m+3)
$$
which has two different solution giving the basis solutions $x^{5/2}$ and $x^{-3/2}$.
Translating back,
$$
u(x)=Ax^{-3/2}+Bx^{5/2}
\implies 
y=\frac{-\frac32Ax^{-5/2}+\frac52Bx^{3/2}}{Ax^{-3/2}+Bx^{5/2}}
=\frac{-3A+5Bx^4}{2Ax+2Bx^5}
$$
and all pairs $(A,B)$ on the same line trough the origin give the same solution.
A: Hint: Using the Substitution $$y(x)=\frac{v(x)}{x}$$ we get
$$v'(x)=\frac{-v(x)^2+v(x)+15}{x}$$ so we have to integrate
$$\int\frac{\frac{dv(x)}{dx}}{-v(x)^2+v(x)+15}dx=\int \frac{1}{x}dx$$
