11.4: Ecuaciones diferenciales homogéneas ordinarias de segundo orden
- Page ID
- 131070
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Este no es un curso matemático completo sobre Ecuaciones diferenciales, pero puede ser útil como recordatorio para quienes ya han estudiado Ecuaciones diferenciales, e incluso puede ser suficiente para nuestros propósitos para aquellos que no lo han hecho.
Suponemos que\( y=y(x)\) and \( y^{'}\) denotes \( \dfrac{dy}{dx}\). An ordinary homogenous second-order differential equation is an Equation of the form
\[ ay^{''}+by^{'}+cy=0, \label{11.4.1} \]
y tenemos que encontrar una función\( y(x)\) which satisfies this. It turns out that it is quite easy to do this, although the nature of the solutions depends on whether \( b^{2}\) is less than, equal to or greater than \( 4ac\).
Un primer punto a tener en cuenta es que, si\( y=f(x)\) is a solution, so is \( Af(x)\) - solo trate de sustituir esto en la Ecuación\( \ref{11.4.1}\). If \( y = g(x)\) is another solution, the same is true of \( g\) - i.e.\( Bg(x)\) is also a solution. And you can also easily verify that any linear combination, such as
\[ y=Af(x)+Bg(x), \label{11.4.2} \]
también es una solución.
Ahora Ecuación\( \ref{11.4.1}\) is a second-order Equation - i.e. the highest derivative is a second derivative - and therefore there can be only two arbitrary constants of integration in the solution - and we already have two in Equation \( \ref{11.4.1}\), and consequently there are no further solutions. All we have to do, then, is to find two functions that satisfy the differential Equation.
No tardará mucho en descubrir que las soluciones de la forma\( y=e^{kx}\) satisfy the Equation, because then \( y^{'}=ky\) and \( y^{''}=k^{2}y\), y, si las sustituyen en la Ecuación\( \ref{11.4.1}\), se obtiene
\[ (ak^{2}+bk+c)y=0. \label{11.4.3} \]
Siempre puedes encontrar dos valores de\( k\) that satisfy this, although these may be complex, which is why the nature of the solutions depends on whether \( b^{2}\) is less than or greater than \( 4ac\). Thus the general solution is
\[ y=Ae^{k_{1}x}+Be^{k_{2}x} \label{11.4.4} \]
donde\( k_{1}\) and \( k_{2}\) are the solutions of the Equation
\[ ax^{2}+bx+c=0. \label{11.4.5} \]
Hay una complicación, sin embargo, si\( b^{2}=4ac\) because then the two solutions of Equation \( \ref{11.4.5}\) are each equal to \(( \dfrac{-b}{(2a)}\). La solución de la Ecuación diferencial es entonces
\[ y=(A+B)\exp\left[\dfrac{-bx}{(2a)}\right] \label{11.4.6} \]
y las dos constantes se pueden combinar en una sola constante\( C=A+B\) so that Equation \( \ref{11.4.6}\) can be written
\[ y=C\exp\left[\dfrac{-bx}{(2a)}\right]. \label{11.4.7} \]
Esta solución solo tiene una constante arbitraria independiente, por lo que debe ser posible una solución adicional. Vamos a tratar de ver si una función de la forma
\[ y=xe^{mx} \label{11.4.8} \]
podría ser una solución de Ecuación\( \ref{11.4.1}\). From Equation \( \ref{11.4.8}\) we obtain \( y^{'}=(1+mx)e^{mx}\) and \( y^{''}=m(2+mx)e^{mx}\). Al sustituir estos en Ecuación\( \ref{11.4.1}\), recordando que\( c=\dfrac{b^{2}}{(4a)}\), obtenemos para el lado izquierdo de la ecuación\( \ref{11.4.1}\), después de algo de álgebra,
\[ \dfrac{e^{mx}}{4a}[(2am+b)^{2}x+4a(2am+b)]. \label{11.4.9} \]
Esto es idénticamente cero si\( m=\dfrac{-b}{(2a)}\), and hence
\[ y=x\exp\left[\dfrac{-bx}{(2a)}\right] \label{11.4.10} \]
es una solución de Ecuación\( \ref{11.4.1}\) the general solution of Equation \( \ref{11.4.1}\), if \( b^{2}=4ac\), is therefore
\[ y=(C=Dx)\exp\left[\dfrac{-bx}{(2a)}\right].\label{11.4.11} \]
Descubriremos cómo son realmente estas soluciones en la siguiente sección.