Author Topic: TUT0401 quiz2 solution  (Read 317 times)

EroSkulled

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TUT0401 quiz2 solution
« on: October 04, 2019, 02:00:01 PM »
Solve$ \frac{x}{(x^2+y^2)^\frac{3}{2}} + \frac{y}{(x^2+y^2)^\frac{3}{2}}\frac{dy}{dx}=0$

Let $M=\frac{x}{(x^2+y^2)^\frac{3}{2}}$ and $N=\frac{y}{(x^2+y^2)^\frac{3}{2}}$
We want to find $N_x$ and $M_y$
\begin{equation}
  M_y=\frac{d}{dy}\frac{x}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}
\end{equation}
\begin{equation}
  N_x=\frac{d}{dx}\frac{y}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}
\end{equation}
\begin{equation}
  N_x=M_y
\end{equation}
We conclude the given equation is exact. We integrate to find $\varphi(x,y)$
Note $\varphi_x(x,y)=M$, integrate both side, we get:
\begin{equation}
  \varphi(x,y)=\int{Mdx}=\int{\frac{x}{(x^2+y^2)^\frac{3}{2}}dx}
\end{equation}
Using u subsitution with $u=x^2+y^2$, we get:
\begin{equation}
  \varphi(x,y)=\frac{1}{2}\int{\frac{1}{u^\frac{3}{2}}du}=-u^{-\frac{1}{2}}=-(x^2+y^2)^{-\frac{1}{2}}+h(y)
\end{equation}
Note $\varphi_y(x,y)\equiv N$, then:
\begin{equation}
  \varphi_y(x,y)=\frac{d}{dy}(-(x^2+y^2)^{-\frac{1}{2}}+h(y))
\end{equation}
\begin{equation}
  \varphi_y(x,y)=\frac{y}{(x^2+y^2)^\frac{3}{2}}+h'(y)\equiv \frac{y}{(x^2+y^2)^\frac{3}{2}}
\end{equation}
Hence we can conlude $h'(y)=0$
Then $h(y)=C$
\begin{equation}
  \varphi(x,y)=-(x^2+y^2)^{-\frac{1}{2}}+C
\end{equation}
The implicit solutoin is $-(x^2+y^2)^{-\frac{1}{2}}=C$