<s>[INST] What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$? I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I have to integrate it. I would think that the $-x$ would equal to $\frac{-x^2}{2}$ but the answer on the back of my book is: $\displaystyle \frac{y^2}{2} - \frac{2}{5} y^{5/2} =\frac{x^3}{3}-x-\frac{54}{5}$
What is the particular solution given $y=4$ and $x=3$?
My Solution
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$
 [/INST] The ODE is $$\displaystyle \small xy\frac{dy}{dx} = \frac{x^3-x}{1- \sqrt y},$$ which can be rewritten as
$$\displaystyle \small (y-y^{3/2}) \ dy  = (x^2-1) \ dx. \ $$ This is the step where you have a mistake! By integrating both sides we get:
$$\displaystyle \small \frac{y^2}{2} - \frac{2}{5} y^{5/2} = \frac{x^3}{3} - x + C,$$
Now, plug in $\small y = 4, x = 3$ and so you'll get $\small C = \displaystyle - \frac{54}{5}$.
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