# Second-Order Linear Homogeneous Differential Equations with Constant Coefficients

#### physics

So I am learning these particular Differential Equation, and in the book (Schaum's 8th Edition Differential Equations) I have quoted an excerpt that I do not understand for clarification. I believe my professor is busy so any help would be great.

It walks you through the three cases of 2nd order linear homogenous DE's (Distinct, real roots, repeated roots, and complex roots) and says the following:

Warning: The above solutions are not valid if the differential equation is not linear or does not have constant coefficients. Consider, for example, the equation y″ – x2y = 0. The roots of the characteristic equation are λ1 = x and λ2 = –x, but the solution is not The reason I ask this is because y″ – x2y = 0 seems like a linear differential equation. Basically what I am trying to say is how do I determine if a linear DE has constant coefficients or variable coefficients? Because looking at the just mentioned function, it seems like it is applicable to this case.

#### Archie

It is linear but does not have constant coefficients. The solutions you have been reading about appky only to equations that are both 1) linear; and 2) have constant coefficients.

#### HallsofIvy

MHF Helper
Warning: The above solutions are not valid if the differential equation is not linear or does not have constant coefficients.
As Archie said, this differential equation does NOT have "constant coefficients". Specifically, the coefficient of y is $-x^2$, not a constant.

#### physics

Is there any way to tell? I get it from context clues that it is has undetermined coefficients but just move on for now? It is has not been told yet how to tell if the coefficients are constant or variable?

#### physics

From what I gather from reading, homogenous differential equations of 2nd order or higher will have constant coefficients. Where they are nonhomogenous is where they will have undetermined coefficients. 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.

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.

#### Archie

Is there any way to tell? I get it from context clues that it is has undetermined coefficients but just move on for now? It is has not been told yet how to tell if the coefficients are constant or variable?
Do you not understand the terms "constant" and "variable"?

• 1 person

#### Archie

From what I gather from reading, homogenous differential equations of 2nd order or higher will have constant coefficients. Where they are nonhomogenous is where they will have undetermined coefficients. 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.

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.
Homogeneous means that there are no terms that do not contain $y$ or one of it's derivatives. The Wronskian and linear independence have nothing to do with it.

#### HallsofIvy

MHF Helper
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.