MAT244-2018S > Quiz-1

TUT 0601

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Darren Zhang:
Question
Find the general solution of the given function.
$$\frac{dy}{dx} = -\frac{(4x+3y)}{(2x+y)}$$

Answer
Let y = xv,
$\frac{dy}{dx} = v + v'x$
$v+v'x = -\frac{4x+3xv}{2x+xv}$
$v+v'x = -\frac{(4+3v)}{(2+v)}$
$v'x = \frac{-v^2-5v-4}{2+v}$
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{x} dx$
Let u = $(v^2+5v+4)$, du = (2v+5)dv
Then,
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{2} \frac{(2v+5)}{(v^2+5v+4)}dv - \int \frac{0.5}{(v^2+5v+4)}dv$
$\frac{1}{2} ln(v^2+5v+4)dv - \frac{1}{2}*\frac{1}{3} \int(\frac{1}{v+1}-\frac{1}{v+4})dv$
$3ln(v^2+5v+4)-ln(\frac{(v+1)}{(v+4)}) = c-6ln(x)$
$(v+1)^2(v+1)^4 = x^6$
Therefore, we can get to know that
$(v+4)^2(v+1) = \frac{c}{x^{3}}$
plug y = xv into this equation
Therefore, we can get
$$(4x+y)^{2} (x+y) = c$$

Meng Wu:
$${dy\over dx}=-{4x+3y\over 2x+y}$$
If we divide right side of the given equation by $x$, we get:
$${dy\over dx}=-{4+3{y\over x} \over 2+{y\over x}}$$
Hence, given equation is a Homogeneous.$\\$
Let $v={y\over x}$, $v$ is a function of $x$.$\\$
Thus, $${dy\over dx}=-{4+3v\over 2+v}$$
Since $v={y\over x} \rightarrow y=vx$   $\\$
By differentiating both sides with respect to $x$, we get:
$${dy\over dx}=v+x{dv\over dx}=-{4+3v\over 2+v}$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-v$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-{v(2+v)\over 2+v}$$
$$x{dv\over dx}={-(v^2+5v+4)\over 2+v}$$
$${dv\over dx}={-(v+1)(v+4)\over 2+v}\cdot {1\over x}$$
$${2+v\over -(v+1)(v+4)}\cdot {dv\over dx}={1\over x}$$
$$-{1\over x}+{2+v\over -(v+1)(v+4)}\cdot {dv\over dx}=0$$
$${1\over x}+{2+v\over (v+1)(v+4)}\cdot {dv\over dx}=0$$
Notice that above equation has the form of $M(x)+N(y){dy\over dx}=0$ $\\$
Hence the equation is Seperable. $\\$
Rewrite the equation, we get:
$${2+v\over (v+1)(v+4)}dv=-{1\over x}dx$$
Integrating both sides, we get:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{-{1\over x}dx}$$
For $\int{{2+v\over (v+1)(v+4)}dv}$, we use Partial Fraction:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{({A\over v+1}+{B\over v+4})dv}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{A(v+4)+B(v+1)\over (v+1)(v+4)}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{(A+B)v+(4A+B)\over (v+1)(v+4)}$$
Hence, $$\begin{cases} A+B=1 \\4A+B=2\\ \end{cases} \implies \begin{cases} A={1\over3}\\ B={2\over3} \end{cases}$$
Thus, rewrite gets us:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{{1\over 3(v+1)}+{2\over 3(v+4)}dv}=-\int{1\over x}dx$$
$${1\over 3}ln|v+1|+{2\over3}ln|v+4|=-ln|x|+c$$
where c is arbitrary constant. $\\$
Multiply both sides by $3$, we get:
$$ln|v+1|+ln|v+4|^2=-3ln|x|+3c$$
Now substitute $v={y\over x}$ back in:
$$ln|{y\over x}+1|+ln|{y\over x}+4|^2=-3ln|x|+3c$$
$$ln|{y+x\over x}|+ln|{y+4x\over x}|^2=-3ln|x|+3c$$
$$ln|y+x|-ln|x|+2ln|y+4x|-2ln|x|=-3ln|x|+3c$$
$$ln|y+x|+ln|y+4x|^2=3c$$
$$ln|y+x||y+4x|^2=3c$$
$$exp^{ln|y+x||y+4x|^2}=exp^{3c}$$
$$|y+x||y+4x|^2=c$$
where c is another arbitrary constant.
Therefore, we get the solution to the given differential equation.

Victor Ivrii:
Meng Wu
This cow can be milked only once

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