Author Topic: QUIZ 1 TUT0402  (Read 516 times)

kaye

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QUIZ 1 TUT0402
« on: September 27, 2019, 02:00:09 PM »
Question: Find the general solution of the given function.
$$\frac{dy}{dx}=-\frac{(4x+3y)}{(2x+y)}$$

On RHS, divide both the numerator and the denominator by $x$:
$$\frac{dy}{dx} = -\frac{(4x+3y)}{(2x+y)}
= -\frac{4\frac{x}{x}+3\frac{y}{x}}{2\frac{x}{x}+\frac{y}{x}}$$

Let $u = \frac{x}{y}$, so $y = ux$, then on LHS we have
$$\frac{dy}{dx}
= \frac{d(ux)}{dx}
= \frac{du}{dx}x + u$$

and on RHS we have
$$-\frac{4\frac{x}{x}+3\frac{y}{x}}{2\frac{x}{x}+\frac{y}{x}}
= -\frac{(4+3u)}{(2+u)}$$

put them together:
\begin{align}
    \frac{du}{dx}x+u & = -\frac{(4+3u)}{(2+u)}\notag\\
    \frac{du}{dx}x & = -\frac{4+3u+2u+u^2}{2+u}\notag\\
    & = -\frac{u^2+5u+4}{2+u}\notag\\
    & = -\frac{(u+1)(u+4)}{2+u}\notag
\end{align}

rearrange this equation:
\begin{align}
    -\frac{2+u}{(u+1)(u+4)}du &=\frac{1}{x}dx\notag\\
    -\int\frac{2+u}{(u+1)(u+4)}du &=\int\frac{1}{x}dx\notag\\
    -\int\frac{1}{3}\frac{1}{u+1}+\frac{2}{3}\frac{1}{u+4}du & =\int\frac{1}{x}dx\notag\\
    \frac{1}{3}\ln{|u+1|}+\frac{2}{3}\ln{|u+4|}&=-\ln{|x|}+C\notag\\
    \ln{|u+1|}+2\ln{|u+4|} &=-3\ln{|x|}+3C\notag\\
    \ln{|\frac{y}{x}+1|}+2\ln{|\frac{y}{x}+4|} &=-3\ln{|x|}+3C\notag\\
    \ln{|\frac{y+x}{x}|}+2\ln{|\frac{y+4x}{x}|} &=-3\ln{|x|}+3C\notag\\
    \ln{|y+x|}-\ln{|x|}+2(\ln{|y+4x|}-\ln{|x|}) &=-3\ln{|x|}+3C\notag\\
    \ln{|y+x|}+2\ln{|y+4x|} &=3C\notag\\
    e^{\ln{|y+x|}}+e^{2\ln{|y+4x|}} &=e^{3C}\notag\\
    |y+x||y+4x|^2 &=e^{3C}\notag
\end{align}

So, the general solution is $|y+x||y+4x|^2=C$.