Non-homogeneous Higher Order LODEs with cc.

class: center, middle, inverse, title-slide

.title[
# Non-homogeneous Higher Order LODEs with cc.
]
.subtitle[
## Session 04
]
.author[
### Alejandro Ucan
]
.date[
### 2023-09-24
]

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# Goals

* Understand how is made a general solution of a non-homogenous LODEs with cc. 
* Describe the function models and the relation with the non-homogenous term. 
* Describe the method of undetermined coefficients. 
* Apply the undetermined coefficients method to solve non-homogenous LODEs with cc.

---
# Recall:

> A non-homogeneous LODEs with cc is of the form `$$a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_2y''+a_1y'+a_0y=g(x),$$` where `$a_n,a_{n-1},\cdots,a_2,a_1,a_0$` are constants and `$g(x)$` is a function.

The particular solution for a non-homogeneous ODE is made by the sum of the general solution of the homogeneous ODE and a particular solution of the non-homogeneous ODE.

From now, we will focus on non-homogeneous LODEs with cc therefore we will omit the "non-homogeneous" word.

---
## Some assumptions:

> The function `$g(x)$` could be any type of function, but we will only consider the following ones: 
 * Polynomials (including constants) `$$g(x)=c_0+c_1x+c_2x^2+\cdots+c_nx^n.$$`
 * Exponentials: `$$g(x)=e^{cx}$$`
 * Sines or Cosines: `$$g(x)=\sin(cx)\mbox{ or } g(x)=\cos(cx).$$`
 * Combinations of these (product and/or sums): `$g(x)=10,$` `$g(x)=15x-6+8e^{-x}$` o `$g(x)=xe^x \sin(x)+(3x^2-1)e^{-4x}.$`
 
---
## The method of undetermined coefficients:

> The method is applied for non-homogeneous LODEs with cc and the non-homogeneous term is a function `$g(x)$` as in the list above.
 
The idea behind this method: "the derivatives of functions `$g(x)$` as in the list above, are combinations of functions in the list.
---
## The method of undetermined coefficients:

1. Find the general solution of the homogeneous LODEs with cc. 
1. For the particular solution, find a model function `$y_p(x)$` and substitute it into the LODEs with cc. 
1. From the fact that we are assuming that is a solution, it must satisfies the LODEs with cc. 
1. The substitution will lead us to a system of equations for the coefficients of the model function.

---
#### Example 1:

Find the particular solution of `$y''+4y'-2y=2x^2-3x+6.$` 
--
 
#### Solution Example 1:
1. From the fact that `$g(x)$` is a polynomial of degree two, then the particular solution could be a polynomial of degree two (equal), let's assume that is `$$y_p=Ax^2+Bx+C.$$`

1. Substituting `$y_p$` in the LODEs with cc: `$y_p'=2Ax+B$` and `$y_p''=2A$` then `$$2A+8Ax+4B-2Ax^2-2Bx-2C=2x^2-3x+6.$$`

1. Solving the system of equations we get `$A=-1,$` `$B=-\frac{5}{2}$` and `$C=-9.$`

1. Therefore `$y_p=-x^2-\frac{5}{2}x-9.$`

---
#### Example 2:

Find the general solution to `$y''-y'+y=2\sin(3x).$` 
--
 
#### Solution Example 1:
1. From the fact that `$g(x)=2\sin(3x)$` is a sine function, then the particular solution could be a sine function but the derivatives of `$\sin(x)$` involve `$\cos$`, let's assume that is `$$y_p=A\sin(3x)+B\cos(3x).$$`

1. Substituting `$y_p$` in the LODEs with cc: `$y_p'=3A\cos(3x)-3B\sin(3x)$` and `$y_p''=-9A\sin(3x)-9B\cos(3x)$` then `$$-9A\sin(3x)-9B\cos(3x)-3A\cos(3x)+3B\sin(3x)+A\sin(3x)+B\cos(3x)=2\sin(3x).$$`
---
#### Example 2:

1. We obtain the following system of equations: `$$\begin{cases} -8A+6B=0 \\ 6A+8B=2 \end{cases}.$$` Solving the system we get `$A=\frac{3}{10}$` and `$B=\frac{1}{10}.$`

1. Therefore the particular solution is `$y_p=\frac{3}{10}\sin(3x)+\frac{1}{10}\cos(3x).$`

1. The general solution is `$y=y_h+y_p=C_1e^x+C_2e^x+\frac{3}{10}\sin(3x)+\frac{1}{10}\cos(3x).$`
---
## An error in the matrix

Find the solution for `$$y''-5y'+4y=8e^x.$$` 
--
 
Now try with the following model function: `$$y_p=Axe^x.$$`

---
## Example 3:

Find the particular solution to `$y''-2y'-3y=4x-5+6xe^{2x}.$`

In this case the particular solution should be of the form: `$$y_p=Ax+B+Cxe^{2x}+Ee^{2x}.$$`