Higher Order Linear Differential Equations with constant coefficients

class: center, middle, inverse, title-slide

.title[
# Higher Order Linear Differential Equations with constant coefficients
]
.subtitle[
## Session 03
]
.author[
### Alejandro Ucan
]
.date[
### 2023-09-24
]

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# Goals

* Describe the general form of a linear homogeneous differential equation with constant coefficients and the IVP. 
* Determine the general solution for a linear homogeneous differential equation with constant coefficients. 
* Deduce the method to solve linear homogeneous differential equations with constant coefficients. 
* Apply the method to solve linear homogeneous differential equations with constant coefficients.

---
# Linear homogeneous differential equations with constant coefficients.

> __Definition:__ the generic form of a linear homogeneous differential equation of order `$n$` is: `$$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots+a_2(x)y''+a_1(x)y'+a_0(x)y=g(x).$$`

To define an IVP we need to add the initial conditions: `$$y(x_0)=y_0,\, y'(x_0)=y_1,\,y''(x_0)=y_2,\cdots,\,y^{(n-1)}(x_0)=y_{n-1}.$$`

---
#### Example 1:

> For the `$3y'''+7y''-7y'-3y=0$` we have that `$y(x)=ce^{x}$` and `$y(x)=ce^{-x/3}+ke^{-3x}$` are solutions.

--

> The function `$y=3e^{2x}+e^{-2x}-3x$` is the solution for the IVP `$y''-4y=12x$` with initial conditions `$y(0)=4$` and `$y'(0)=1.$`

---
# First Order LODE with cc.

> The general form of a first order LODE with cc is `$ay'+by=g(x),$` lets try to solve the homogeneous case. 
 * The solution have to be of the form `$y=e^{mx}.$` 
 * Substituting in the LODE we have that `$$ame^{mx}+be^{mx}=0=e^{mx}(am+b),$$` 
 * From the fact that `$e^{mx}\neq 0$` we have that `$$am+b=0\Rightarrow m=-b/a.$$`
 * Therefore the solution is `$y(x)=ce^{-bx/a}.$`
---
# Second Order LODE with cc.

> Lets take the second order homogeneous LODE with cc. 
 * The solution have to be of the form `$y=e^{mx}.$` 
 * Substituting in the LODE we obtain that our __auxiliary polynomial__ is `$$am^2+bm+c=0.$$` 
 * We have that there two roots to this polynomio (they might be egual): `$m_1$` and `$m_2.$` 
 * Remember that if:
 * If `$\Delta=b^2-4ac>0$` then the roots are real and different.
 * If `$\Delta=0$` then the roots are real and equal.
 * If `$\Delta<0$` then the roots are complex conjugates.
 
---
## First Case: Different real roots:

> In the case the roots are real and different, `$m_1$` and `$m_2,$` then the general solution is `$$y=c_1 e^{m_1 x}+c_2e^{m_2 x}.$$`

--
 
#### Example 1:

Find the general solution to `$2y''-5y'-3y=0,$`

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#### Solution:
* The auxiliary polynomio is `$2m^2-5m-3=0$` 
* The roots are `$m_1=3$` and `$m_2=-1/2$` 
* Therefore, the general solution is: `$$y=c_1e^{3x}+c_2e^{-x/2}.$$`
---
## Second Case: Equal real roots:

> In the case the roots are real and equal, `$m_1=m_2=m,$` then the general solution is `$$y=c_1 e^{mx}+c_2xe^{mx}.$$`

--
 
#### Example 2:
Find the general solution to `$y''-10y'+25y=0.$`

--
 
#### Solution:
* The auxiliary polynomio is `$m^2-10m+25=0$` 
* The roots are `$m_1=m_2=5$` 
* Therefore, the general solution is: `$$y=c_1e^{5x}+c_2xe^{5x}.$$`

---
## Third Case: Complex conjugate roots:
> In the case the roots are complex conjugates, `$m_1=a+bi$` and `$m_2=a-bi,$` then the general solution is `$$y=c_1 e^{a}\cos(bx)+c_2e^{ax}\sin(bx).$$`

--
 
#### Example 3:
Find the general solution to `$y''+4y'+7y=0.$`

--
 
#### Solution:
* The auxiliary polynomio is `$m^2+4m+7=0$` 
* The roots are `$m_1=-2+\sqrt{3}i$` and `$m_2=-2-\sqrt{3}i.$` 
* Therefore, the general solution is: `$$y=c_1e^{-2x}\cos(\sqrt{3}x)+c_2e^{-2x}\sin(\sqrt{3}x).$$`
---
#### Example 4:

Find the solution of `$4y''+4y'+17y=0$` if `$y(0)=-1$` y `$y'(0)=2.$`

--
 
#### Solution:
* The auxiliary polynomio is `$4m^2+4m+17=0$` 
* The roots are `$m_1=-\frac{1}{2}-2i$` and `$m_2=-\frac{1}{2}+2i.$` 
* Therefore, the general solution is: `$$y=c_1e^{-\frac{1}{2}x}\cos(2x)+c_2e^{-\frac{1}{2}x}\sin(2x).$$`
* Applying the initial condition we have the following system of equations: `$$\begin{cases} c_1=-1 \\ -\frac{1}{2}c_1+2c_2=2 \end{cases}$$`
whose solution are `$c_1=-1$` and `$c_2=1.$` Therefore the solution is `$$y=-e^{-\frac{1}{2}x}\cos(2x)+e^{-\frac{1}{2}x}\sin(2x).$$`