What is a good way/ways (in "lay" terms) to know if an equation is separable or linear, especially for cases where it is not as obvious from first glance?
Tips on recognizing Separable /Linear Equations
1 Answers
Here is what comes in my mind:
Consider the following differential equation:
$$y'=\frac{dy}{dx}=\frac{ax+by+c}{a'x+b'y+c'}.$$
If $\det \left(\begin{array}{cc} a&b \\ a'&b' \end{array}\right)=0,$ then the change of variables $T=ax+by$ or $T=a'x+b'y$ where $\frac{dT}{dx}=a+by'$ or $\frac{dT}{dx}=a'+b'y'$ makes it into a seperable diff equation.
If $\det \left(\begin{array}{cc} a&b \\ a'&b' \end{array}\right)\not=0,$ then we should find the solutions of
$$\left\{\begin{array}{cc}ax+by+c=0 \\ a'x+b'y+c'=0 \end{array}\right.$$
namely, $x=\alpha, y=\beta.$ By translation of the origin to the intersection of the above lines, we may write
$$\left\{\begin{array}{cc}x=X+\alpha \\ y=Y+\beta \end{array}\right.$$
where $\frac{dy}{dx}=\frac{dY}{dX}.$ Therefore, by this changing of variables, we will have
$$Y'=\frac{dY}{dX}=\frac{aX+bY}{a'X+b'Y}$$
and this can be solved by a new change of variables $T(x)=y/x$ that will make it into a separable diff equation.
Exercises:
1) $y'=\frac{1}{x+y}.$
2) $(3x+2y-4)dx+(x+y+2)dy=0.$
-
0Who said the equation was of that form? My reading of the question is that it's asking when an equation _is_ separable or linear, not when it can be transformed into one that is separable or linear. – 2011-07-11
-
0Yes, when given an equation, since I understand how to use the exact method, I almost immediately go for that method of solving the D.E. And this I do because I have a hard time telling if an equation is separable or not. – 2011-07-11