# Solving Rational ODE's of the form (ax+by+c) dx+(ex+fy+g) dy=0.

• MHB
• Ackbach
In summary, there are three cases for the rational ODE $(ax+by+c)\,dx+(ex+fy+g)\,dy=0,$ based on the two straight lines multiplying the differentials. The first case is when the two lines are identical, in which case there is a constant, non-zero ratio of the coefficients. The second case is when the lines are parallel and non-intersecting, with a non-zero ratio of the coefficients. The third case is when the lines intersect at one point, which can be determined algebraically. Each case requires a different approach to solve the ODE.
Ackbach
Gold Member
MHB
There are essentially three cases of the rational ODE $(ax+by+c)\,dx+(ex+fy+g)\,dy=0,$ since there are two straight line expressions multiplying the differentials. We will think of this geometrically, then translate to the algebraic approach. The tricky part to these problems is keeping track of all the sub-cases. I will usually stop before getting too much into the sub-cases; the idea is that you should be able to take it from where I lead you.

Case 1. 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 $k:$
\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.

Case 2. 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
\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.

Case 3: 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:
\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 $y.$

Summary. 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.

Well, that concludes this note on Differential Equations!

Hello there,

Thank you for your thorough explanation of the three cases for the rational ODE. It's clear that you have a strong understanding of the subject and I appreciate you breaking down the algebraic and geometric perspectives. Your explanation of how to recognize each case and the steps to take in solving them is very helpful. I will definitely keep your summary in mind when working on similar problems. Thank you for sharing your knowledge and expertise on this topic. Keep up the great work!

## 1. What is a rational ODE?

A rational ODE (ordinary differential equation) is a type of differential equation that can be expressed as a ratio of two polynomials, where the dependent variable and its derivatives appear in the numerator and denominator.

## 2. How do you solve a rational ODE?

To solve a rational ODE, you can use the method of separation of variables, where you isolate the terms containing the dependent variable and its derivatives on one side of the equation and the remaining terms on the other side. This allows you to integrate both sides and find the general solution.

## 3. What is the form of a rational ODE?

The general form of a rational ODE is (ax+by+c)dx + (ex+fy+g)dy = 0, where a, b, c, e, f, and g are constants and x and y are the independent and dependent variables, respectively.

## 4. Can all rational ODEs be solved analytically?

No, not all rational ODEs can be solved analytically. Some may require numerical methods or approximations to find a solution.

## 5. What are some real-world applications of solving rational ODEs?

Rational ODEs are commonly used in physics, engineering, and economics to model various systems and phenomena such as population growth, chemical reactions, and electrical circuits. They can also be used in data analysis and machine learning to make predictions and classify data.

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