# Method of undetermined coefficients

In summary: It's the term that includes both x and f'(x).In summary, the conversation discusses solving a differential equation using the complementary solution and finding a particular solution using the product rule. There is confusion about differentiating terms where the x-factor is not differentiated in the product rule, but it is explained that these terms will disappear and not affect the solution.
hello everyone, i need some huge help here. here's the equation :

y''+y'-6y=10e^2x-18e^3x-6x-11.

complementary solution:c1e^2x+c2e^-3x
s1={e^2x}
s2={e^3x}
s3=(x,1}
ok since e^2x exists in the complimentary solution, it is therefore a solution, so i multiply it by x to get s'1{xe^2x}, so now i have that and my s2 and s3.

so i end up with this:

yp=Axe^2x+Be^3x+Cx+D

heres where i get confused do i use the product rule for Axe^2x, if so do i end up with these two derivatives: y'p=Axe^2x+2Ae^2x+...etc, y''p=Axe^2x+4Ae^2x+...etc. i really get confused with this product rule thing.

"heres where i get confused do i use the product rule for Axe^2x, if so do i end up with these two derivatives: y'p=Axe^2x+2Ae^2x+...etc, y''p=Axe^2x+4Ae^2x+...etc. i really get confused with this product rule thing."

Certainly you are to differentiate the whole thing!
Those derivatives of xe^2x that doesn't differentiate the x-factor, but effectively only differentiates the exponential, will disappear.

why doesn't it differentiate the x factor?

h(x)=f(x)g(x)
h'(x)=f(x)g'(x)+(f'(x)g(x))
h''(x)=f(x)g''(x)+(2f'(x)g'(x)+f''(x))
The paranthesized terms differentiaties f, the other terms regard f effectively as a constant multiple of the g-function.
Assuming some diff.eq:
$$\alpha{h}''+\beta{h}'+\gamma{h}=L(x)$$
we may write this as:
$$f(x)(\alpha{g}''+\beta{g}'+\gamma{g})+(\alpha(2f'(x)g'(x)+f''(x)g)+\beta{f}'g)=L(x)$$
Thus, if g is a solution of the associated homogenous problem, the first term vanishes identically.

Last edited:
arildno said:
Those derivatives of xe^2x that don't differentiate the x-factor, but effectively only differentiates the exponential, will disappear.
You misinterpreted this. arildno meant "in those terms where, using the product rule, the x itself was not differentiated".
For example, differentiating xf(x) using the product rule, I get f(x)+ xf '(x). The term "xf '(x)" is the one arildno was referring to.

## What is the method of undetermined coefficients?

The method of undetermined coefficients is a mathematical technique used to solve nonhomogeneous linear differential equations. It involves finding a particular solution to the equation by assuming a form for the solution and then solving for the coefficients.

## When is the method of undetermined coefficients used?

This method is used to solve nonhomogeneous linear differential equations with constant coefficients, where the nonhomogeneous term is a polynomial, exponential, sine, cosine, or a combination of these functions.

## How does the method of undetermined coefficients work?

The method of undetermined coefficients works by assuming a form for the particular solution based on the type of nonhomogeneous term present in the equation. The coefficients in this assumed solution are then determined by substituting the solution into the original equation and solving for the unknown coefficients.

## What are the advantages of using the method of undetermined coefficients?

One advantage of this method is that it is relatively simple and straightforward to use, especially for equations with simple nonhomogeneous terms. It also allows for the efficient and systematic solution of nonhomogeneous linear differential equations.

## Are there any limitations to the method of undetermined coefficients?

The method of undetermined coefficients can only be used for a limited set of nonhomogeneous linear differential equations. It is not applicable for equations with non-constant coefficients or non-linear terms. In these cases, other methods such as variation of parameters or Laplace transforms may be used.

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