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Home / Differential Equations / Higher Order Differential Equations / Undetermined Coefficients
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Section 7-3 : Undetermined Coefficients

We now need to start looking into determining a particular solution for $$n$$th order differential equations. The two methods that we’ll be looking at are the same as those that we looked at in the 2nd order chapter.

In this section we’ll look at the method of Undetermined Coefficients and this will be a fairly short section. With one small extension, which we’ll see in the lone example in this section, the method is identical to what we saw back when we were looking at undetermined coefficients in the 2nd order differential equations chapter.

Given the differential equation,

${y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = g\left( t \right)$

if $$g\left( t \right)$$ is an exponential function, polynomial, sine, cosine, sum/difference of one of these and/or a product of one of these then we guess the form of a particular solution using the same guidelines that we used in the 2nd order material. We then plug the guess into the differential equation, simplify and set the coefficients equal to solve for the constants.

The one thing that we need to recall is that we first need the complementary solution prior to making our guess for a particular solution. If any term in our guess is in the complementary solution then we need to multiply the portion of our guess that contains that term by a $$t$$. This is where the one extension to the method comes into play. With a 2nd order differential equation the most we’d ever need to multiply by is $${t^2}$$. With higher order differential equations this may need to be more than $${t^2}$$.

The work involved here is almost identical to the work we’ve already done and in fact it isn’t even that much more difficult unless the guess is particularly messy and that makes for more mess when we take the derivatives and solve for the coefficients. Because there isn’t much difference in the work here we’re only going to do a single example in this section illustrating the extension. So, let’s take a look at the lone example we’re going to do here.

Example 1 Solve the following differential equation. ${y^{\left( 3 \right)}} - 12y'' + 48y' - 64y = 12 - 32{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{4t}}$
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We first need the complementary solution so the characteristic equation is,

${r^3} - 12{r^2} + 48r - 64 = {\left( {r - 4} \right)^3} = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}r = 4\,\,\left( {{\mbox{multiplicity 3}}} \right)$

We’ve got a single root of multiplicity 3 so the complementary solution is,

${y_c}\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}t{{\bf{e}}^{4t}} + {c_3}{t^2}{{\bf{e}}^{4t}}$

Now, our first guess for a particular solution is,

${Y_P} = A + B{{\bf{e}}^{ - 8t}} + C{{\bf{e}}^{4t}}$

Notice that the last term in our guess is in the complementary solution so we’ll need to add one at least one $$t$$ to the third term in our guess. Also notice that multiplying the third term by either $$t$$ or $${t^2}$$ will result in a new term that is still in the complementary solution and so we’ll need to multiply the third term by $${t^3}$$ in order to get a term that is not contained in the complementary solution.

Our final guess is then,

${Y_P} = A + B{{\bf{e}}^{ - 8t}} + C{t^3}{{\bf{e}}^{4t}}$

Now all we need to do is take three derivatives of this, plug this into the differential equation and simplify to get (we’ll leave it to you to verify the work here),

$- 64A - 1728B{{\bf{e}}^{ - 8t}} + 6C{{\bf{e}}^{4t}} = 12 - 32{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{4t}}$

Setting coefficients equal and solving gives,

\begin{aligned}{t^0} & : & - 64A &= 12\\{{\bf{e}}^{ - 8t}} & : & - 1728B & = - 32\\{{\bf{e}}^{4t}} & : & 6C & = 2\end{aligned}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{array}{*{20}{l}}{A = - \frac{3}{{16}}}\\{B = \frac{1}{{54}}}\\{C = \frac{1}{3}}\end{array}

A particular solution is then,

${Y_P} = - \frac{3}{{16}} + \frac{1}{{54}}{{\bf{e}}^{ - 8t}} + \frac{1}{3}{t^3}{{\bf{e}}^{4t}}$

The general solution to this differential equation is then,

$y\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}t{{\bf{e}}^{4t}} + {c_3}{t^2}{{\bf{e}}^{4t}} - \frac{3}{{16}} + \frac{1}{{54}}{{\bf{e}}^{ - 8t}} + \frac{1}{3}{t^3}{{\bf{e}}^{4t}}$

Okay, we’ve only worked one example here, but remember that we mentioned earlier that with the exception of the extension to the method that we used in this example the work here is identical to work we did the 2nd order material.