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- Jan 26, 2012

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*The two lines $ax+by+c=0$ and $ex+fy+g=0$ are identical (that is, they intersect at all points on the lines). Algebraically, then, we must have a constant, non-zero ratio of all the coefficients, which we'll call $k:$*

**Case 1.**\begin{align*}

a&=ke\\

b&=kf\\

c&=kg.

\end{align*}

Indeed, this is how you can recognize this case! We can re-write the ODE as

$$k\,(ex+fy+g)\,dx+(ex+fy+g)\,dy=(ex+fy+g)(k\,dx+dy)=0. $$

We can see from inspection that the straight line $ex+fy+g=0$ is a solution of the ODE, but it might not be the only one. Let us suppose that we seek other solutions (so we'll assume, for the moment, that $ex+fy+g\not=0.$) In that case, we can simply cancel the factor and obtain

\begin{align*}

k\,dx+dy&=0 \\

-k\int dx&=\int dy \\

-kx+C&=y,

\end{align*}

which is the equation of another line. So we see that there are two solutions in this case:

$$ex+fy+g=0\quad\text{or}\quad y=-kx+C, $$

where $k$ is defined as above, in terms of $a, b, c, e, f, g.$ If you have an initial value problem, and your initial value is not on the line $ex+fy+g=0,$ then you will have to choose the other solution in order to take advantage of the arbitrary constant. And that wraps up this case fairly well.

**The two lines $ax+by+c=0$ and $ex+fy+g=0$ are parallel and non-intersecting. Algebraically, we must have a non-zero $k$ such that**

*Case 2.*\begin{align*}

a&=ke\\

b&=kf\\

c&\not=kg.

\end{align*}

We can rewrite the DE as

\begin{align*}

(ax+by+c)\,dx+(ex+fy+g)\,dy&=0 \\

(kex+kfy+c)\,dx+(ex+fy+g)\,dy&=0 \\

k(ex+fy+c/k)\,dx+(ex+fy+g)\,dy&=0.

\end{align*}

Now we let $u=ex+fy,$ with $du=e\,dx+f\,dy$ and $du-e\,dx=f\,dy,$ as before. Substituting into the DE yields

\begin{align*}

k(u+c/k)\,dx+(u+g)(du-e\,dx)&=0\\

k\,u\,dx+c\,dx+u\,du-e\,u\,dx+g\,du-eg\,dx&=0\\

(ku+c-eu-eg)\,dx+(u+g)\,du&=0\\

(u(k-e)+c-eg)\,dx+(u+g)\,du&=0.

\end{align*}

From here, you can see that the DE is separable! There are way too many subcases and possibilities to consider, so I will stop here for Case 2.

**The two lines $ax+by+c=0$ and $ex+fy+g=0$ intersect at one point, call it $(x_0,y_0).$ Where is that point? We solve the two equations simultaneously:**

*Case 3:*\begin{align*}

ax_0+by_0&=-c\\

ex_0+fy_0&=-g\\

x_0&=\frac{\left|\begin{matrix}-c&b\\-g&f\end{matrix}\right|}{\left|\begin{matrix}a&b\\e&f\end{matrix}\right|}\\

&=\frac{bg-cf}{af-be}\\

y_0&=\frac{\left|\begin{matrix}a&-c\\e&-g\end{matrix}\right|}{\left|\begin{matrix}a&b\\e&f\end{matrix}\right|}\\

&=\frac{ce-ag}{af-be}.

\end{align*}

Naturally, this is only possible if $af-be\not=0,$ the condition for intersecting lines in the first place.

The idea here is to substitute a translation $t=x-x_0$ and $s=y-y_0.$ This would mean $dt=dx$ and $ds=dy,$ so that the original system becomes

\begin{align*}

\left(a\left(t+x_0\right)+b\left(s+y_0\right)+c\right)dt+\left(e\left(t+x_0\right)+f\left(s+y_0\right)+g\right)ds&=0\\

\left(a\left(t+\frac{bg-cf}{af-be}\right)+b\left(s+\frac{ce-ag}{af-be}\right)+c\right)dt+\left(e\left(t+\frac{bg-cf}{af-be}\right)+f\left(s+\frac{ce-ag}{af-be}\right)+g\right)ds&=0.

\end{align*}

Let's take a look at the constant terms for each differential (so, ignoring $t$ and $s$ for now):

\begin{align*}

a\,\frac{bg-cf}{af-be}+b\,\frac{ce-ag}{af-be}+c&=a\,\frac{bg-cf}{af-be}+b\,\frac{ce-ag}{af-be}+c\,\frac{af-be}{af-be}\\

&=\frac{abg-acf+bce-abg+acf-bce}{af-be}\\

&=0.

\end{align*}

Similarly,

\begin{align*}

e\,\frac{bg-cf}{af-be}+f\,\frac{ce-ag}{af-be}+g&=e\,\frac{bg-cf}{af-be}+f\,\frac{ce-ag}{af-be}+g\,\frac{af-be}{af-be}\\

&=\frac{beg-cef+cef-afg+afg-beg}{af-be}\\

&=0.

\end{align*}

So the constant terms all cancel! That makes the DE as follows:

$$(at+bs)\,dt+(et+fs)\,ds=0.$$

This is a homogeneous ODE, because $t$ and $s$ appear in the same powers. We perform the substitution $tu=s,$ with $ds=t\,du+u\,dt,$ and the DE becomes

\begin{align*}

(at+btu)\,dt+(et+ftu)(t\,du+u\,dt)&=0\\

(at+btu)\,dt+(et+ftu)\,t\,du+(et+ftu)\,u\,dt&=0\\

\left(at+btu+etu+ftu^2\right)dt+\left(et^2+ft^2u\right)du&=0\\

t\left(a+bu+eu+fu^2\right)dt+t^2(e+fu)\,du&=0\\

\left(a+bu+eu+fu^2\right)dt+t(e+fu)\,du&=0\\

\left(a+bu+eu+fu^2\right)dt&=-t(e+fu)\,du\\

\frac{dt}{t}&=-\frac{e+fu}{a+bu+eu+fu^2}\,du\\

\ln|t|+C&=-\int\frac{e+fu}{a+bu+eu+fu^2}\,du.

\end{align*}

This you can solve implicitly for $t.$ Alternatively, you could have done the substitution $t=us,$ with $dt=u\,ds+s\,du,$ which would probably have given you an easier solution for the independent variable. I'll stop here, as I've reduced the problem to an integral with too many sub-cases. But don't forget that once you integrate, you need to back-substitute, first to get back to $s$ and $t,$ and then un-translate to get to $x$ and $y.$

**If the two lines $ax+by+c=0$ and $ex+fy+g=0$ are identical, then substitute $u=ax+by+c.$ If they are parallel and non-intersecting, substitute $u=ax+by.$ If they intersect at one point $(x_0,y_0),$ then substitute $t=x-x_0$ and $s=y-y_0$ to lead to a homogeneous equation, where either $tu=s$ or $t=su$ will render the equation separable.**

*Summary.*Well, that concludes this note on Differential Equations!