Author Topic: tut0402 question  (Read 319 times)

shangluy

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tut0402 question
« on: October 04, 2019, 02:00:04 PM »
Solve \begin{align*}
    \frac{x dx}{(x^2 + y^2)^\frac{3}{2}} + \frac{y dy}{(x^2 + y^2)^\frac{3}{2}} = 0
\end{align*}

We can multiple both side by $(x^2 + y^2)^\frac{3}{2}$, then we have
\begin{equation}
    xdx + ydy = 0
\end{equation}

Let $M = x$ and $N = y$
\begin{equation}
    M_y = 0\quad \quad N_x = 0
\end{equation}

Since $M_y = N_x$,  the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
    \psi(x, y) &= \int M dx \\
    &= \int x dx\\
    &= \frac{1}{2}x^2 + h(y)\\
\end{align*}
Since
\begin{align*}
    \frac{\partial \psi}{\psi y} &= h'(y)\\
    \implies h'(y) &= y\\
    \implies h(y) &= \frac{1}{2}y^2 + C
\end{align*}
Therefore, the solution is $\frac{1}{2}x^2 + \frac{1}{2}y^2 = C$, rearrange we get $x^2 + y^2 = C$, where $x \neq 0$ and $y \neq 0$