<s>[INST] Problem in Differential Equation. Find particular solution when x=1 and y=1
MY attempts, 
$$xy^2\frac{dy}{dx}-x^3-y^3=0$$
Solving as homogeneous equation, 
$$-x^3-y^3+xy^2\frac{dy}{dx}=0$$
Let $$y=xv$$
$$\frac{dy}{dx}=v+x\frac{dy}{dx}$$
$$-x^3-x^3v^3+x^3(x\frac{dv}{dx}+v)v^2=0$$
$$x^3(x\frac{dv}{dx}v^2-1)=0$$
$$\frac{dv}{dx}=\frac{1}{xv^2}$$
$$\frac{dv}{dx}v^2=\frac{1}{x}$$
$$\int \frac{dv}{dx}v^2dx=\int \frac{1}{x}dx$$
$$\frac{v^3}{3}=\ln (x)+c_1$$
$$v=\sqrt[3]{3}\sqrt[3]{ln (x)+c_1}$$
$$y=xv$$
$$y=x\sqrt[3]{3\ln (x)+c_1}$$
When y=1, x=1
$c_1=1$$
Therefore, $$y=x\sqrt[3]{3\ln (x)+1}$$
But the given answer is $v^3=\ln (ex^3)$. I wonder why. Can anybody explain?
 [/INST] $$\begin{array}{l}
x{y^2}\frac{{dy}}{{dx}} - {x^3} - {y^3} = 0 \Rightarrow \frac{{dy}}{{dx}} - \frac{{{x^2}}}{{{y^2}}} - \frac{y}{x} = 0\\
u = \frac{y}{x} \Rightarrow xu = y \Rightarrow udx + xdu = dy\\
\frac{{xdu}}{{dx}} - \frac{1}{{{u^2}}} = 0 \Rightarrow {u^2}du = \frac{{dx}}{x}\\
\frac{{{u^3}}}{3} = \ln (x) + \ln ({{c^{ - 3}}}) \Rightarrow {u^3} = 3\ln ({{c^{ - 3}}}x) = \ln (c{x^3})\\
{u^3} = \ln (c{x^3})\\
{\left( {\frac{y}{x}} \right)^3} = \ln (c{x^3})\\
x = 1 \wedge y = 1 \Rightarrow \ln (c) = 1 \Rightarrow c = e\\
{\left( {\frac{y}{x}} \right)^3} = \ln (e{x^3})
\end{array}$$
They didn't solve it for $y$.
</s>