Lecture5 - method of undetermined coefficients#

This method gives us guide to find the nonhomogeneous (particular) solutions. Consider the general form of second order ODE:

\[\begin{eqnarray*} A''+by'+cy = q(t) \end{eqnarray*}\]

We guess the form of nonhomogeneous (particular) solutions based on the form of q(t):

\[\begin{eqnarray*} q(t) &\rightarrow& y_{p}(t)\\\\ t, 1 &\rightarrow& a+bt\\\\ e^{st} &\rightarrow& Ye^{st} \text{(if resonance: } Yt^{n}e^{st})\\\\ te^{st} &\rightarrow& (a+bt)e^{st}\\\\ \sin{}(\omega t), \cos{}(\omega t) &\rightarrow& M\cos{}(\omega t) + N\sin{}(\omega t)\\\\ t\cos{}(\omega t) &\rightarrow& (a+bt)\cos{}(\omega t) + (c+dt)\sin{}(\omega t)\\\\ e^{i\omega t}, e^{-i\omega t} &\rightarrow& Ye^{-i\omega t} \end{eqnarray*}\]

Example 6

\[\begin{eqnarray*} y''+5y'+6y=e^{4t} \end{eqnarray*}\]

  1. homogeneous solution:

\[\begin{eqnarray*} r^2+5r+6=0 \rightarrow r = -2, -3\\\\ \Longrightarrow y_{h}(t) = c_{1}e^{-2t}+c_{2}e^{-3t} \end{eqnarray*}\]

  1. nonhomogeneous/particular solution by guessing the form of source/input term. We have one undetermined coefficient Y.

\[\begin{eqnarray*} y_{p}(t) &=& Ye^{4t} \rightarrow 16Y+20Y+6Y = 1 \rightarrow Y=\frac{1}{42} \\\\ &\Longrightarrow& y_{p}(t) = \frac{1}{42}e^{4t} \end{eqnarray*}\]

  1. general/complete solution:

\[\begin{eqnarray*} y(t) = c_{1}e^{-2t}+c_{2}e^{-3t} + \frac{1}{42}e^{4t} \end{eqnarray*}\]

Example 7

\[\begin{eqnarray*} y''+5y'=e^{it} \end{eqnarray*}\]

  1. homogeneous solution:

\[\begin{eqnarray*} r^2+r=0 \rightarrow r(r+1)=0 \rightarrow = 0, -1\\\\ \Longrightarrow y_{h}(t) = c_{1}e^{0t}+c_{2}e^{-t} = c_{1}+c_{2}e^{-t} \end{eqnarray*}\]

  1. nonhomogeneous/particular solution by guessing the form of source/input term. We have one undetermined coefficient Y.

\[\begin{eqnarray*} y_{p}(t) &=& Ye^{it} \rightarrow Y(i^{2}+i) = 1 \rightarrow Y=\frac{1}{i-1} \\\\ &\Longrightarrow& y_{p}(t) = \frac{1}{i-1}e^{it} \end{eqnarray*}\]

  1. general/complete solution:

\[\begin{eqnarray*} y(t) = c_{1}+c_{2}e^{-t} + \frac{1}{i-1}e^{it} \end{eqnarray*}\]

Example 8

\[\begin{eqnarray*} y'''+2y''+y'=e^{3t} \end{eqnarray*}\]

  1. homogeneous solution:

\[\begin{eqnarray*} r^{3}+2r^{2}+r=0 \rightarrow r(r+1)^{2}=0 \rightarrow = 0, -1, -1\\\\ \Longrightarrow y_{h}(t) = c_{1}e^{0t}+c_{2}e^{-t} + c_{3}te^{-t} = c_{1}+c_{2}e^{-t} + c_{3}te^{-t} \end{eqnarray*}\]

  1. nonhomogeneous/particular solution by guessing the form of source/input term. We have one undetermined coefficient Y.

\[\begin{eqnarray*} y_{p}(t) &=& Ye^{3t} \rightarrow 27Y+18Y+3Y = 1 \rightarrow Y=\frac{1}{48} \\\\ &\Longrightarrow& y_{p}(t) = \frac{1}{i-1}e^{it} \end{eqnarray*}\]

  1. general/complete solution:

\[\begin{eqnarray*} y(t) = c_{1}+c_{2}e^{-t} + c_{3}te^{-t} + \frac{1}{48}e^{3t} \end{eqnarray*}\]

Example 9

\[\begin{eqnarray*} y''+y'+y=t \end{eqnarray*}\]

  • nonhomogeneous/particular solution by guessing the form of source/input term. We have two undetermined coefficients.

\[\begin{eqnarray*} y_{p}(t) &=& a+bt \rightarrow b+a+bt = t \rightarrow b=1, a=-1 \\\\ &\Longrightarrow& y_{p}(t) = -1+t \end{eqnarray*}\]

Example 10

\[\begin{eqnarray*} y''+y'+y=\sin{}(t) \end{eqnarray*}\]

  • nonhomogeneous/particular solution by guessing the form of source/input term. We have two undetermined coefficient Y.

\[\begin{eqnarray*} y_{p}(t) &=& c_{1}\cos{}(t) + c_{2}\sin{}(t) \\\\ &\Longrightarrow& -c_{1}\cos{}(t) - c_{2}\sin{}(t) - c_{1}\sin{}(t) + c_{2}\cos{}(t) + c_{1}\cos{}(t) + c_{2}\sin{}(t) = \sin{}(t) \\\\ &\Longrightarrow& c_{2} = 0, c_{1}=-1 \\\\ &\Longrightarrow& y_{p}(t) = -\cos{}(t) \end{eqnarray*}\]

Example 11

\[\begin{eqnarray*} y''+y'+y=t\sin{}(t) \end{eqnarray*}\]

  • nonhomogeneous/particular solution by guessing the form of source/input term. We have four undetermined coefficients.

\[\begin{eqnarray*} y_{p}(t) &=& (a+bt)\cos{}(t) + (c+dt)\sin{}(t) \end{eqnarray*}\]