Author Topic: TUT0702 Quiz1  (Read 571 times)

Yue Sagawa

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TUT0702 Quiz1
« on: September 27, 2019, 03:38:53 PM »
Question: \begin{equation*} \frac{dy}{dx} = \frac{x^2-3y^2}{2xy} \end{equation*}
  Show that the given equation is homogeneous. Solve the differential equation.
  The equation can be written as:
\begin{equation*} \frac{dy}{dx} = \frac{1-3 (\frac{y}{x})^2}{2(\frac{y}{x})}\end{equation*}
It can be expressed as a function of \begin{equation*}\frac{y}{x}\end{equation*} So the equation is homogeneous.
Let \begin{equation*} u = \frac{y}{x} \end{equation*}Then \begin{equation*} y = ux \end{equation*} and
\begin{equation*} \frac{dy}{dx} = u + x\frac{dv}{dx}\end{equation*}
Substitute, we get:
\begin{equation*} u + x\frac{dv}{dx} = \frac{1-3u^2}{2u}\end{equation*}
\begin{equation*} x\frac{dv}{dx} = \frac{1-3u^2}{2u} - u \end{equation*}
Observe the equation is separable.
\begin{equation*} \int_{}{} \frac{2u}{1-5u^2} du = \int_{}{} \frac{1}{x} dx \end{equation*}
\begin{equation*} -\frac{1}{5} ln|1-5u^2| = ln|x| + c_1\end{equation*}
\begin{equation*}- ln|1-5u^2| = 5 ln|x| + c\end{equation*}
Substitute back,
\begin{equation*}- ln|1-5\frac{y^2}{x^2}| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}- ln|\frac{x^2-5y^2}{x^2}| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|+ln|x^2| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|+2ln|x| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|-3ln|x| = c_2\end{equation*}
\begin{equation*}-ln|x^3(x^2-5y^2)| = c_3\end{equation*}
\begin{equation*}|x^3(x^2-5y^2) = c\end{equation*}
\begin{equation*}|x^3||x^2-5y^2| = c\end{equation*} where c is an arbitrary constant.