Este método es otra alternativa con respecto al método de coeficientes indeterminados para encontrar una solución particular de la ecuación
\[\begin{equation}p\left( t \right)y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\label{eq:eq1}\end{equation}\]
y vimos que si bien reducía las cosas a un problema de álgebra, el álgebra podía volverse bastante desordenado. Además de eso, los coeficientes indeterminados solo funcionarán para una clase de funciones bastante pequeña.
El método de variación de parámetros es un método mucho más general que puede usarse en muchos más casos. Sin embargo, hay dos desventajas en el método. Primero, la solución complementaria es absolutamente necesaria para resolver el problema. Esto contrasta con el método de coeficientes indeterminados, en el que era aconsejable tener a mano la solución complementaria pero no era necesario. Segundo, como veremos, para completar el método, haremos un par de integrales y no hay garantía de que podamos hacer las integrales. Entonces, si bien siempre será posible escribir una fórmula para obtener la solución particular, es posible que no podamos encontrarla realmente si las integrales son demasiado difíciles o si no podemos encontrar la solución complementaria.
Vamos a derivar la fórmula para la variación de parámetros. Comenzaremos reconociendo que la solución complementaria para \(\eqref{eq:eq1}\) es
\[{y_c}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)\]
Recuerde también que esta es la solución general a la ecuación diferencial homogénea.
\[\begin{equation}p\left( t \right)y'' + q\left( t \right)y' + r\left( t \right)y = 0\label{eq:eq2}\end{equation}\]
Recuerde también que para anotar la solución complementaria sabemos que \(y_{1}(t)\) and \(y_{2}(t)\) Son un conjunto fundamental de soluciones.
Lo que vamos a hacer es ver si podemos encontrar un par de funciones, \(u_{1}(t)\) y \(u_{2}(t)\) así que
\[{Y_P}\left( t \right) = u{_1}\left( t \right){y_1}\left( t \right) + u{_2}\left( t \right){y_2}\left( t \right)\]
Será una solución para \(\eqref{eq:eq1}\). Tenemos dos incógnitas aquí y, por lo tanto, necesitaremos dos ecuaciones eventualmente. Una ecuación es fácil. Nuestra solución propuesta debe satisfacer la ecuación diferencial, por lo que obtendremos la primera ecuación al conectar nuestra solución propuesta en \(\eqref{eq:eq1}\). La segunda ecuación puede provenir de una variedad de lugares. Vamos a obtener nuestra segunda ecuación simplemente haciendo una suposición que facilitará nuestro trabajo. Vamos a decir más sobre esto en breve.
Entonces, vamos a empezar. Si vamos a insertar nuestra solución propuesta en la ecuación diferencial, vamos a necesitar algunos derivados, así que vamos a obtenerlos. La primera derivada es \[{Y'_P}\left( t \right) = {u'_1}{y_1} + {u_1}{y'_1} + {u'_2}{y_2} + {u_2}{y'_2}\]
Aquí está el supuesto. Simplemente para hacer que la primera derivada sea más fácil de tratar, vamos a asumir que cualquiera que sea \(u_{1}(t)\) y \(u_{2}(t)\) Son ellos van a satisfacer los siguientes
\[\begin{equation}{u'_1}{y_1} + {u'_2}{y_2} = 0\label{eq:eq3}\end{equation}\]
Ahora, no hay razón de antemano para creer que esto se puede hacer. Sin embargo, veremos que esto funcionará. Simplemente hacemos esta suposición con la esperanza de que no causará problemas en el futuro y para hacer que la primera derivada sea más fácil, así que no se emocione.
Con este supuesto se convierte la primera derivada.
\[{Y'_P}\left( t \right) = {u_1}{y'_1} + {u_2}{y'_2}\]
La segunda derivada es entonces, \[{Y''_P}\left( t \right) = {u'_1}{y'_1} + {u_1}{y''_1} + {u'_2}{y'_2} + {u_2}{y''_2}\]
Plug the solution and its derivatives into \(\eqref{eq:eq1}\).
\[p\left( t \right)\left( {{{u'}_1}{{y'}_1} + {u_1}{{y''}_1} + {{u'}_2}{{y'}_2} + {u_2}{{y''}_2}} \right) + q\left( t \right)\left( {{u_1}{{y'}_1} + {u_2}{{y'}_2}} \right) + r\left( t \right)\left( {u{_1}{y_1} + u{_2}{y_2}} \right) = g\left( t \right)\]
Rearranging a little gives the following.
\[\begin{align*}& p\left( t \right)\left( {{{u'}_1}{{y'}_1} + {{u'}_2}{{y'}_2}} \right) + {u_1}\left( t \right)\left( {p\left( t \right){{y''}_1} + q\left( t \right){{y'}_1} + r\left( t \right){y_1}} \right) + \\ & \hspace{3.0in}{u_2}\left( t \right)\left( {p\left( t \right){{y''}_2} + q\left( t \right){{y'}_2} + r\left( t \right){y_2}} \right) = g\left( t \right)\end{align*}\]
Now, both \(y_{1}(t)\) and \(y_{2}(t)\) are solutions to \(\eqref{eq:eq2}\) and so the second and third terms are zero. Acknowledging this and rearranging a little gives us,
\[p\left( t \right)\left( {{{u'}_1}{{y'}_1} + {{u'}_2}{{y'}_2}} \right) + {u_1}\left( t \right)\left( 0 \right) + {u_2}\left( t \right)\left( 0 \right) = g\left( t \right)\]
\[\begin{equation}{u'_1}{y'_1} + {u'_2}{y'_2} = \frac{{g\left( t \right)}}{{p\left( t \right)}}\label{eq:eq4}\end{equation}\]
We’ve almost got the two equations that we need. Before proceeding we’re going to go back and make a further assumption. The last equation, \(\eqref{eq:eq4}\), is actually the one that we want, however, in order to make things simpler for us we are going to assume that the function \(p(t) = 1\).
In other words, we are going to go back and start working with the differential equation,
\[y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\]
If the coefficient of the second derivative isn’t one divide it out so that it becomes a one. The formula that we’re going to be getting will assume this! Upon doing this the two equations that we want to solve for the unknown functions are
\[\begin{equation}{u'_1}{y_1} + {u'_2}{y_2} = 0\label{eq:eq5}\end{equation}\]
\[\begin{equation}{u'_1}{y'_1} + {u'_2}{y'_2} = g\left( t \right)\label{eq:eq6}\end{equation}\]
Note that in this system we know the two solutions and so the only two unknowns here are \({u'_1}\) and \({u'_2}\). Solving this system is actually quite simple. First, solve \(\eqref{eq:eq5}\) for \({u'_1}\) and plug this into \(\eqref{eq:eq6}\) and do some simplification.
\[\begin{equation}{u'_1} = - \frac{{{{u'}_2}{y_2}}}{{{y_1}}}\label{eq:eq7}\end{equation}\]
\[\begin{align*}\left( { - \frac{{{{u'}_2}{y_2}}}{{{y_1}}}} \right){{y'}_1} + {{u'}_2}{{y'}_2} & = g\left( t \right)\\ {{u'}_2}\left( {{{y'}_2} - \frac{{{y_2}{{y'}_1}}}{{{y_1}}}} \right) & = g\left( t \right)\\ {{u'}_2}\left( {\frac{{{y_1}{{y'}_2} - {y_2}{{y'}_1}}}{{{y_1}}}} \right) & = g\left( t \right)\end{align*}\]
\[\begin{equation}{u'_2} = \frac{{{y_1}g\left( t \right)}}{{{y_1}{{y'}_2} - {y_2}{{y'}_1}}}\label{eq:eq8}\end{equation}\]
So, we now have an expression for \({u'_2}\). Plugging this into \(\eqref{eq:eq7}\) will give us an expression for \({u'_1}\).
\[\begin{equation}{u'_1} = - \frac{{{y_2}g\left( t \right)}}{{{y_1}{{y'}_2} - {y_2}{{y'}_1}}}\label{eq:eq9}\end{equation}\]
Next, let’s notice that
\[W\left( {{y_1},{y_2}} \right) = {y_1}{y'_2} - {y_2}{y'_1} \ne 0\]
Recall that \(y_{1}(t)\) and \(y_{2}(t)\) are a fundamental set of solutions and so we know that the Wronskian won’t be zero!
Finally, all that we need to do is integrate \(\eqref{eq:eq8}\) and \(\eqref{eq:eq9}\) in order to determine what \(u_{1}(t)\) and \(u_{2}(t)\) are. Doing this gives,
\[{u_1}\left( t \right) = - \int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{u_2}\left( t \right) = \int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\]
So, provided we can do these integrals, a particular solution to the differential equation is
\[\begin{align*}{Y_P}\left( t \right) & = {y_1}{u_1} + {y_2}{u_2}\\ & = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\end{align*}\]
So, let’s summarize up what we’ve determined here.
Variation of Parameters
Consider the differential equation,
\[y'' + q\left( t \right)y' + r\left( t \right)y = g\left( t \right)\]
Assume that \(y_{1}(t)\) and \(y_{2}(t)\) are a fundamental set of solutions for
\[y'' + q\left( t \right)y' + r\left( t \right)y = 0\]
Then a particular solution to the nonhomogeneous differential equation is,
\[{Y_P}\left( t \right) = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}}\]
Depending on the person and the problem, some will find the formula easier to memorize and use, while others will find the process used to get the formula easier. The examples in this section will be done using the formula.
Before proceeding with a couple of examples let’s first address the issues involving the constants of integration that will arise out of the integrals. Putting in the constants of integration will give the following.
\[\begin{align*}{Y_P}\left( t \right) & = - {y_1}\left( {\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + c} \right) + {y_2}\left( {\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + k} \right)\\ & = - {y_1}\int{{\frac{{{y_2}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + {y_2}\int{{\frac{{{y_1}g\left( t \right)}}{{W\left( {{y_1},{y_2}} \right)}}\,dt}} + \left( { - c{y_1} + k{y_2}} \right)\end{align*}\]
The final quantity in the parenthesis is nothing more than the complementary solution with c1 = -c and \(c\)2 = k and we know that if we plug this into the differential equation it will simplify out to zero since it is the solution to the homogeneous differential equation. In other words, these terms add nothing to the particular solution and so we will go ahead and assume that \(c\) = 0 and \(k\) = 0 in all the examples.
One final note before we proceed with examples. Do not worry about which of your two solutions in the complementary solution is \(y_{1}(t)\) and which one is \(y_{2}(t)\). It doesn’t matter. You will get the same answer no matter which one you choose to be \(y_{1}(t)\) and which one you choose to be \(y_{2}(t)\).
Let’s work a couple of examples now.
Find a general solution to the following differential equation.
\[2y'' + 18y = 6\tan \left( {3t} \right)\]
First, since the formula for variation of parameters requires a coefficient of a one in front of the second derivative let’s take care of that before we forget. The differential equation that we’ll actually be solving is
\[y'' + 9y = 3\tan \left( {3t} \right)\]
We’ll leave it to you to verify that the complementary solution for this differential equation is
\[{y_c}\left( t \right) = {c_1}\cos \left( {3t} \right) + {c_2}\sin \left( {3t} \right)\]
So, we have
\[{y_1}\left( t \right) = \cos \left( {3t} \right)\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = \sin \left( {3t} \right)\]
The Wronskian of these two functions is
\[W = \left| {\begin{array}{*{20}{c}}{\cos \left( {3t} \right)}&{\sin \left( {3t} \right)}\\{ - 3\sin \left( {3t} \right)}&{3\cos \left( {3t} \right)}\end{array}} \right| = 3{\cos ^2}\left( {3t} \right) + 3{\sin ^2}\left( {3t} \right) = 3\]
The particular solution is then,
\[\begin{align*}{Y_P}\left( t \right) & = - \cos \left( {3t} \right)\int{{\frac{{3\sin \left( {3t} \right)\tan \left( {3t} \right)}}{3}\,dt}} + \sin \left( {3t} \right)\int{{\frac{{3\cos \left( {3t} \right)\tan \left( {3t} \right)}}{3}\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\frac{{{{\sin }^2}\left( {3t} \right)}}{{\cos \left( {3t} \right)}}\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\frac{{1 - {{\cos }^2}\left( {3t} \right)}}{{\cos \left( {3t} \right)}}\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \cos \left( {3t} \right)\int{{\sec \left( {3t} \right) -
\cos \left( {3t} \right)\,dt}} + \sin \left( {3t} \right)\int{{\sin \left( {3t} \right)\,dt}}\\ & = - \frac{{\cos \left( {3t} \right)}}{3}\left( {\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right| - \sin \left( {3t} \right)} \right) + \frac{{\sin \left( {3t} \right)}}{3}\left( { - \cos \left( {3t} \right)} \right)\\ & = - \frac{{\cos \left( {3t} \right)}}{3}\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right|\end{align*}\]
The general solution is,
\[y\left( t \right) = {c_1}\cos \left( {3t} \right) + {c_2}\sin \left( {3t} \right) - \frac{{\cos \left( {3t} \right)}}{3}\ln \left| {\sec \left( {3t} \right) + \tan \left( {3t} \right)} \right|\]
Find a general solution to the following differential equation.
\[y'' - 2y' + y = \frac{{{{\bf{e}}^t}}}{{{t^2} + 1}}\]
We first need the complementary solution for this differential equation. We’ll leave it to you to verify that the complementary solution is,
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}t{{\bf{e}}^t}\]
So, we have
\[{y_1}\left( t \right) = {{\bf{e}}^t}\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = t{{\bf{e}}^t}\]
The Wronskian of these two functions is
\[W = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^t}}&{t{{\bf{e}}^t}}\\{{{\bf{e}}^t}}&{{{\bf{e}}^t} + t{{\bf{e}}^t}}\end{array}} \right| = {{\bf{e}}^t}\left( {{{\bf{e}}^t} + t{{\bf{e}}^t}} \right) - {{\bf{e}}^t}\left( {t{{\bf{e}}^t}} \right) = {{\bf{e}}^{2t}}\]
The particular solution is then,
\[\begin{align*}{Y_P}\left( t \right) & = - {{\bf{e}}^t}\int{{\frac{{t{{\bf{e}}^t}\,{{\bf{e}}^t}}}{{{{\bf{e}}^{2t}}\left( {{t^2} + 1} \right)}}\,dt}} + t{{\bf{e}}^t}\int{{\frac{{{{\bf{e}}^t}\,{{\bf{e}}^t}}}{{{{\bf{e}}^{2t}}\left( {{t^2} + 1} \right)}}\,dt}}\\ & = - {{\bf{e}}^t}\int{{\frac{t}{{{t^2} + 1}}\,dt}} + t{{\bf{e}}^t}\int{{\frac{1}{{{t^2} + 1}}\,dt}}\\ & = - \frac{1}{2}{{\bf{e}}^t}\ln \left( {1 + {t^2}} \right) + t{{\bf{e}}^t}{\tan ^{ - 1}}\left( t
\right)\end{align*}\]
The general solution is,
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}t{{\bf{e}}^t} - \frac{1}{2}{{\bf{e}}^t}\ln \left( {1 + {t^2}} \right) + t{{\bf{e}}^t}{\tan ^{ - 1}}\left( t \right)\]
This method can also be used on non-constant coefficient differential equations, provided we know a fundamental set of solutions for the associated homogeneous differential equation.
Find the general solution to
\[ty'' - \left( {t + 1} \right)y' + y = {t^2}\]
given that
\[{y_1}\left( t \right) = {{\bf{e}}^t}\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = t + 1\]
form a fundamental set of solutions for the homogeneous differential equation.
As with the first example, we first need to divide out by a \(t\).
\[y'' - \left( {1 + \frac{1}{t}} \right)y' + \frac{1}{t}y = t\]
The Wronskian for the fundamental set of solutions is
\[W = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^t}}&{t + 1}\\{{{\bf{e}}^t}}&1\end{array}} \right| = {{\bf{e}}^t} - {{\bf{e}}^t}\left( {t + 1} \right) = - t{{\bf{e}}^t}\]
The particular solution is.
\[\begin{align*}{Y_P}\left( t \right) & = - {{\bf{e}}^t}\int{{\frac{{\left( {t + 1} \right)t}}{{ - t{{\bf{e}}^t}}}\,dt}} + \left( {t + 1} \right)\int{{\frac{{{{\bf{e}}^t}\left( t \right)}}{{ - t{{\bf{e}}^t}}}\,dt}}\\ & = {{\bf{e}}^t}\int{{\left( {t + 1} \right){{\bf{e}}^{ - t}}\,dt}} - \left( {t + 1} \right)\int{{\,dt}}\\ & = {{\bf{e}}^t}\left( { - {{\bf{e}}^{ - t}}\left( {t + 2} \right)} \right) - \left( {t + 1} \right)t\\ & = - {t^2} - 2t - 2\end{align*}\]
The general solution for this differential equation is.
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2t - 2\]
We need to address one more topic about the solution to the previous example. The solution can be simplified down somewhat if we do the following.
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2t - 2\\ & = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2} - 2\left( {t + 1} \right)\\ & = {c_1}{{\bf{e}}^t} + \left( {{c_2} - 2} \right)\left( {t + 1} \right) - {t^2}\end{align*}\]
Now, since \({c_2}\) is an unknown constant subtracting 2 from it won’t change that fact. So we can just write the \({c_2} - 2\) as \({c_2}\) and be done with it. Here is a simplified version of the solution for this example.
\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}\left( {t + 1} \right) - {t^2}\]
This isn’t always possible to do, but when it is you can simplify future work.
