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I have been reading R. Kent Nagle's Fundamental's of Differential Equations textbook and I'm really confused as to the meaning of the terms of "Particular Solution" and "General Solution", specifically as they change from being used in a first-order equation to a second-order equation.

So in a first-order differential equation, your answer should have a "C" somewhere resulting from integrating somewhere. This would be called your general solution because you haven't specified any initial conditions. If you did, and you incorporated that information into your solution, then it would be called your particular solution. Alright, so far so good.

But when I started learning about second-order differential equations, I got really confused because I understood that if you solve a homogeneous second-order linear differential equation with no initial condition information then you would get a general solution with two "C"s because of its second order. This made sense, but when I learned about the Method of Undetermined Coefficients and the Variation of Parameters, I learned that the general solution to a non-homogeneous second-order linear differential equation involved a particular solution AND the general solution to the homogeneous diff eq. But I don't see why or how this could make sense. Does this occur because the terms particular and general are redefined for second-order diff EQs? I'm overall just very confused about the terminology. General clarification would be greatly appreciated.

Edit 1: Thank you both Professor @Robert Israel and @K.defaoite

It makes a bit more sense, but I'm still overall confused. To be a little more explicit, why does the Method of Undetermined Coefficients give you a particular solution? Again, I'm very used to the idea that a particular solution is a general solution with initial conditions applied to it (from first-order differential equations) but I don't see the particular-ness that the Method of Undetermined Coefficients gives you in the way that solving an initial value problem for a first-order differential equation does. Thanks again to you both.

R. Kent Nagle - Method of Undetermined Coefficients

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  • $\begingroup$ Another word in use for the homogeneous solution is "complementary" solution, so you get "general = complementary + particular = homogeneous + non-homogeneous". $\endgroup$ Commented Aug 4, 2020 at 8:16
  • $\begingroup$ Hey @Moo, I see where you’re coming from but I don’t think I’m quite there yet. So does this mean that the nonhomogeneity, f(t), makes a homogeneous “general” linear diff eq into a non-homogeneous “particular” Diff eq? If so, why? How does the nonhomogeneity specify a particular solution? And why does homogeneous linear differential result in a general solution? $\endgroup$
    – Noah A.
    Commented Aug 4, 2020 at 18:52

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The meaning is the same: a particular solution is just that, one solution, corresponding to one choice of initial conditions. If you have a formula for the general solution, in a second order equation it will have two arbitrary parameters and each choice of values for those parameters gives you a particular solution.

Linear equations (for any order) have the property that if you add a solution of the homogeneous equation and a solution of the non-homogeneous equation, you get another solution of the same non-homogeneous equation. If you take the general solution of the homogeneous equation (involving, for a second-order equation, two parameters), and add a particular solution of the non-homogeneous equation, you get the general solution of the non-homogeneous equation.

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Mathematicians uses the word solution in regard to differential equations somewhat careless. Any function satisfying the equation they call solution and assign it later the adjective particular or general, with or without fulfilled boundary conditions. It seems like there are different kinds of solutions for the same equation. It is very confusing and leads to misconceptions especially in physics (so called nonphysical solutions).

Lets assume a linear second order homogeneous ODE subjected to two physically motivated boundary conditions. If one finds a function $f(x)$ satisfying the ODE then function $y=a f(bx)$, with constants $a$ and $b$ determined by boundary conditions $y(x_1)=y_1$ and $y'(x_1)=y_1'$ is called the solution of the problem. It is however completely wrong. The solution of the second order linear differential equation is a vector in two dimensional solution space. Therefore, it hast to be a linear combination of two linearly independent base functions $f_1(x)$ and $f_2(x)$ that both satisfy the equation. The correct or true solution is then $$ y(x)=C_1 f_1(x)+C_2 f_2(x).$$ The coefficients determined by boundary conditions as $$C_1=\frac{y_1' f_2'(x_1)-y_1 f_2(x_1)}{W},~~~~C_2=-\frac{y_1' f_1'(x_1)-y_1 f_1(x_1)}{W},$$ where $$W\equiv f_1(x_1) f_2'(x_1)-f_2(x_1) f_1'(x_1).$$

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  • $\begingroup$ Given a solution $f$ to a homogeneous ODE $Df=0$ (linear), yes, differentiating $f(bx)$ with respect to $b$, also known as "variation of parameters", is a device that can be used to solve $Dg=f$. This is somewhat separate from the idea that the solution space of a second-order ODE is two-dimensional, and that we need two "boundary conditions" to determine those constants. $\endgroup$ Commented Nov 2, 2023 at 19:38
  • $\begingroup$ @paulgarrett Thanks for your comment. These two ways of solving equation produce different solutions. I have thought that finding a solution to an ODE given an initial value - the initial value problem - results in one unique solution. It should be independent of the solution's way. $\endgroup$
    – JanG
    Commented Nov 3, 2023 at 9:20
  • $\begingroup$ Yes, there are uniqueness results for boundary-value problems... $\endgroup$ Commented Nov 3, 2023 at 15:10

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