From what I gather from reading, homogenous differential equations of 2nd order or higher will have constant coefficients.

No, that's not true at all! I'm not sure whether you are having problems with the word "constant" or "coefficient" but it must be one or the other!

Both equations, $3\frac{d^2y}{d^2x}+ 2\frac{dy}{dx}+ 5y= 0$

**and** $e^x\frac{d^2y}{d^2x}- x^2\frac{dy}{dx}+ cos(x)y= 0$ are "linear homogeneous second order differential equations. The first has coefficients 3, 2, and 5- constants. The second has coefficients $e^x$, $-x^2$, and $cos(x)$- not constants.

Where they are nonhomogenous is where they will have undetermined coefficients

First the phrase "undetermined coefficients" refers to a solution technique, not the differential equation itself. If we had, for example, the differential equation $y''+ 3y+ 2y= e^x+ 3e^{2x}$, the "associated homogenous equation" is $y''+ 3y'+ 2y= 0$. That has "characteristic equation" $r^2+ 3r+ 2= (r+ 2)(r+ 1)= 0$ so has general solution $Ae^{-2x}+ Be^{-x}$.

**Because** the "non-homogeneous part", $e^x+ 3e^{2x}$, has exponentials, the type of solution we would expect for a "linear differential equation with constant coefficients", we try a solution to the entire equation of the form $C_1e^x+ C_2e^{2x}$. The "$C_1$" and "$C_2$" are "undetermined coefficients". Setting $y= C_1e^x+ C_2e^{2x}$, $y'= C_1e^x+ 2C_2e^{2x}$, and $y''= C_1e^x+ 4C_2e^{2x}$. Putting those into the equation, $y''+ 3y'+ 2y= C_1e^x+ 4C_2e^{2x}+ 3C_1e^x+ 6C_2e^{2x}+ 2C_1e^{x}+ 2C_2e^{2x}= 7C_1e^x+ 12e^{2x}= e^x+ 3e^{2x}$ so we must have $7C_1= 1$ and $12C_2= 3$. $C_1= \frac{1}{7}$ and $C_2= \frac{1}{4}$ so the general solution to the entire differential equation is $Ae^{-x}+ Be^{-2x}+ \frac{1}{7}e^x+ \frac{1}{4}e^{2x}$. Again, "undetermined coefficients" is a

**method** for solving non-homogeneous differential equations, not a type of equation. And, it really only works where we can predict the form of the solution which is true only when the "non-homogeneous" part is of a form we would expect for solutions of "linear differential equations"- polynomials, exponential, sine or cosine or products of those. Another method, "variation of parameters", is harder but more general.

I believe how you tell the if it is homogeneous or non-homogenous is in its linear independence. If the Wronskian returns roots it is linearly dependent. If the Wronskian returns anything other than a zero, it is linearly indepedent and nonhomogenous.

No, no, no! First, the terms "homogeneous" and "non-homogeneous" only apply to linear differential equations. A second order linear differential equation is of the form $p(x)\frac{d^2y}{dx^2}+ q(x)\frac{dy}{dx}+ r(x)y= f(x)$. The differential equation is "homogeneous" if and only if $f(x)$ is identically 0.

Like in an equation, y^(4) + 10x^2y * y^(3) + y = g(x)y, you would plug in the functions of x or constants (x^2, 2, 3, etc) into the Wronskian, and test for linear independence or dependence. That will tell you whether it is homogenous or heterogenous, and homogenous higher order differential equations always have constant coefficients. It gets easier as you go along . IF there are any corrections anyone wants to make or parts that need further thought, please contribute as this thread as I believe it is important information.

Unfortunately, pretty much nothing you say here is correct! The equation you give,

**as you write it**, $y^{(4)} + 10x^2y y^{(3)} + y = g(x)y$, could be written $y^{(4)}+ 10x^2y * y^{(3)}+ (1- g(y)) y = 0$

**is** homogeneous because every term is a multiple of y or a derivative of y. I suspect you intended $y^(4) + 10x^2y * y^(3) + y = g(x)$ which is

**non-homogeneous** because the "g(x)" is not multiplied by y or a derivative of y.

And, by the way, because of the "y" multiplying the third derivative this is a

**non-linear** differential equation.