---
title: "NUS-MATH570 Problem Set 1"
source: https://www.jemoka.com/posts/kbhnus_math570_problem_set_1/
---

We have:
\begin{equation} \frac{2y^{2}}{9-x^{2}} + y \dv{y}{x} + \frac{3y}{2-x} = 0 \end{equation}
We want to get rid of things; let&rsquo;s begin by dividing the whole thing by \(y\).
\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} + \frac{3}{2-x} = 0 \end{equation}
Finally, then, moving the right expression to the right, we have:
\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} = \frac{-3}{2-x} \end{equation}
In this case, we have functions:
\begin{equation} \begin{cases} P(x) = \frac{2}{9-x^{2}}\\ Q(x) = \frac{-3}{2-x}\\ \end{cases} \end{equation}
Taking first the top integral:
\begin{equation} \int \frac{2}{9-x^{2}} \dd{x} = \frac{1}{3} \log \qty(\frac{x+3}{3-x}) \end{equation}
Raising \(e\) to that power, we have that:
\begin{equation} \sqrt[3]{e\frac{x+3}{3-x}} \end{equation}
Multiplying \(Q(x)\) to that expression, we have that:
\begin{equation} \int \frac{-3}{2-x}\sqrt[3]{e\cdot \frac{x+3}{3-x}} \dd{x} \end{equation}
Therefore, our entire answer is defined as the integral function that:
\begin{equation} y = \frac{1}{\sqrt[3]{e\cdot \frac{x+3}{3-x}} } \int \frac{-3}{2-x}\sqrt[3]{e\cdot \frac{x+3}{3-x}} \dd{x} \end{equation}
