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
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$

Okay, so you have to solve the equation:
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
The first step is to move things around, which you did properly.
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}$$
Now we can move the $dx$ to the other side and write the equation as,
$$y(1-\sqrt y)dy=\frac{1}{x}\cdot (x^3-x)dx$$
Simplifying and adding in the integration,
$$\int \left(y-y^\frac{3}{2}\right)dy= \int(x^2-1)dx$$
So, integrating both sides, we have,
$$\frac{y^{1+1}}{(1+1)}-\frac{y^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}=\frac{x^{2+1}}{(2+1)}-x+C\\
\frac{1}{2}y^2-\frac{2}{5}y^{\frac{5}{2}}=\frac{1}{3}x^3-x+C$$
Finally, plugging in $y=4$ and $x=3$, we have,
$$\frac{1}{2}(4)^2-\frac{2}{5}(4)^{\frac{5}{2}}=\frac{1}{3}(3)^3-3+C\\
-\frac{24}{5}=6+C\\ C=-\frac{54}{5}$$ or, we can make it a little cleaner by saying that there is some other constant, say $C_1$, such that, $$C_{1}=\frac{54}{5}$$
