Differential Equations - Higher Order: Variation of Parameters - ii

So, we’ve got $n$ equations, but notice that just like we got when we did this for 2^nd order differential equations the unknowns in the system are not $u_1\left(t\right),u_2\left(t\right),\dots ,u_n\left(t\right)$ but instead they are the derivatives, $u_1^{\prime }\left(t\right),u_2^{\prime }\left(t\right),\dots ,u_n^{\prime }\left(t\right)$. This isn’t a major problem however. Provided we can solve this system we can then just integrate the solutions to get the functions that we’re after.

Also, recall that the $y_1\left(t\right),y_2\left(t\right),\dots ,y_n\left(t\right)$ are assumed to be known functions and so they along with their derivatives (which appear in the system) are all known quantities in the system.

Now, we need to think about how to solve this system. If there aren’t too many equations we can just solve it directly if we want to. However, for large $n$ (and it won’t take much to get large here) that could be quite tedious and prone to error and it won’t work at all for general $n$ as we have here.

The best solution method to use at this point is then Cramer’s Rule. We’ve used Cramer’s Rule several times in this course, but the best reference for our purposes here is when we used it when we first defined Fundamental Sets of Solutions back in the 2^nd order material.

Upon using Cramer’s Rule to solve the system the resulting solution for each $u_i^{\prime }$ will be a quotient of two determinants of $n$x $n$ matrices. The denominator of each solution will be the determinant of the matrix of the known coefficients,

$$\left|\begin{array}{cccc}{y}_1& y_2& \cdots & y_n\\ {y^{\prime }}_1& {y^{\prime }}_2& \cdots & {y^{\prime }}_n\\ ⋮& ⋮& \ddots & ⋮\\ y_1^{\left(n-1\right)}& y_2^{\left(n-1\right)}& \cdots & y_n^{\left(n-1\right)}\end{array}\right|=W\left(y_1,y_2,\dots y_n\right)\left(t\right)$$

This however, is just the Wronskian of $y_1\left(t\right),y_2\left(t\right),\dots ,y_n\left(t\right)$ as noted above and because we have assumed that these form a fundamental set of solutions we also know that the Wronskian will not be zero. This in turn tells us that the system above is in fact solvable and all of the assumptions we apparently made out of the blue above did in fact work.

The numerators of the solution for $u_i^{\prime }$ will be the determinant of the matrix of coefficients with the $i$^th column replaced with the column $\left(0,0,0,\dots ,0,g\left(t\right)\right)$. For example, the numerator for the first one, $u_1^{\prime }$ is,

$$\left|\begin{array}{cccc}0& y_2& \cdots & y_n\\ 0& {y^{\prime }}_2& \cdots & {y^{\prime }}_n\\ ⋮& ⋮& \ddots & ⋮\\ g\left(t\right)& y_2^{\left(n-1\right)}& \cdots & y_n^{\left(n-1\right)}\end{array}\right|$$

Now, by a nice property of determinants if we factor something out of one of the columns of a matrix then the determinant of the resulting matrix will be the factor times the determinant of new matrix. In other words, if we factor $g\left(t\right)$ out of this matrix we arrive at,

$$\left|\begin{array}{cccc}0& y_2& \cdots & y_n\\ 0& {y^{\prime }}_2& \cdots & {y^{\prime }}_n\\ ⋮& ⋮& \ddots & ⋮\\ g\left(t\right)& y_2^{\left(n-1\right)}& \cdots & y_n^{\left(n-1\right)}\end{array}\right|=g\left(t\right)\left|\begin{array}{cccc}0& y_2& \cdots & y_n\\ 0& {y^{\prime }}_2& \cdots & {y^{\prime }}_n\\ ⋮& ⋮& \ddots & ⋮\\ 1& y_2^{\left(n-1\right)}& \cdots & y_n^{\left(n-1\right)}\end{array}\right|$$

We did this only for the first one, but we could just as easily done this with any of the $n$ solutions. So, let $W_i$ represent the determinant we get by replacing the $i$^th column of the Wronskian with the column $\left(0,0,0,\dots ,0,1\right)$ and the solution to the system can then be written as,

$$u_1^{\prime }=\frac{g\left(t\right)W_1\left(t\right)}{W\left(t\right)},u_2^{\prime }=\frac{g\left(t\right)W_2\left(t\right)}{W\left(t\right)},\cdots ,u_n^{\prime }=\frac{g\left(t\right)W_n\left(t\right)}{W\left(t\right)}$$

Wow! That was a lot of effort to generate and solve the system but we’re almost there. With the solution to the system in hand we can now integrate each of these terms to determine just what the unknown functions, $u_1\left(t\right),u_2\left(t\right),\dots ,u_n\left(t\right)$we’ve after all along are.

$$u_1=\int \frac{g\left(t\right)W_1\left(t\right)}{W\left(t\right)}dt,u_2=\int \frac{g\left(t\right)W_2\left(t\right)}{W\left(t\right)}dt,\cdots ,u_n=\int \frac{g\left(t\right)W_n\left(t\right)}{W\left(t\right)}dt$$

Finally, a particular solution to $\text{(1)}$ is then given by,

$$Y\left(t\right)=y_1\left(t\right)\int \frac{g\left(t\right)W_1\left(t\right)}{W\left(t\right)}dt+y_2\left(t\right)\int \frac{g\left(t\right)W_2\left(t\right)}{W\left(t\right)}dt+\cdots +y_n\left(t\right)\int \frac{g\left(t\right)W_n\left(t\right)}{W\left(t\right)}dt$$

We should also note that in the derivation process here we assumed that the coefficient of the $y^{\left(n\right)}$ term was a one and that has been factored into the formula above. If the coefficient of this term is not one then we’ll need to make sure and divide it out before trying to use this formula.

Before we work an example here we really should note that while we can write this formula down actually computing these integrals may be all but impossible to do.

Okay let’s take a look at a quick example.

Example 1 Solve the following differential equation.$$y^{\left(3\right)}-2y^{\prime \prime }-21y^{\prime }-18y=3+4e^{-t}$$

The characteristic equation is,
$$r^3-2r^2-21r-18=\left(r+3\right)\left(r+1\right)\left(r-6\right)=0\Rightarrow r_1=-3,r_2=-1,r_3=6$$So, we have three real distinct roots here and so the complimentary solution is,
$$y_c\left(t\right)=c_1{e}^{-3t}+c_2{e}^{-t}+c_3{e}^{6t}$$Okay, we’ve now got several determinants to compute. We’ll leave it to you to verify the following determinant computations.
$$\begin{array}{cccc}W& =\left|\begin{array}{ccc}{e}^{-3t}& e^{-t}& e^{6t}\\ -3e^{-3t}& -e^{-t}& 6e^{6t}\\ 9e^{-3t}& e^{-t}& 36e^{6t}\end{array}\right|=126e^{2t}& W_1& =\left|\begin{array}{ccc}0& e^{-t}& e^{6t}\\ 0& -e^{-t}& 6e^{6t}\\ 1& e^{-t}& 36e^{6t}\end{array}\right|=7e^{5t}\\ W_2& =\left|\begin{array}{ccc}{e}^{-3t}& 0& e^{6t}\\ -3e^{-3t}& 0& 6e^{6t}\\ 9e^{-3t}& 1& 36e^{6t}\end{array}\right|=-9e^{3t}& W_3& =\left|\begin{array}{ccc}{e}^{-3t}& e^{-t}& 0\\ -3e^{-3t}& -e^{-t}& 0\\ 9e^{-3t}& e^{-t}& 1\end{array}\right|=2e^{-4t}\end{array}$$Now, given that $g\left(t\right)=3+4e^{-t}$ we can compute each of the $u_i$. Here are those integrals.
$$u_1=\int \frac{\left(3+4e^{-t}\right)\left(7e^{5t}\right)}{126e^{2t}}dt=\frac{1}{18}\int 3e^{3t}+4e^{2t}dt=\frac{1}{18}\left(e^{3t}+2e^{2t}\right)$$$$u_2=\int \frac{\left(3+4e^{-t}\right)\left(-9e^{3t}\right)}{126e^{2t}}dt=-\frac{1}{14}\int 3e^t+4dt=-\frac{1}{14}\left(3e^t+4t\right)$$$$u_3=\int \frac{\left(3+4e^{-t}\right)\left(2e^{-4t}\right)}{126e^{2t}}dt=\frac{1}{63}\int 3e^{-6t}+4e^{-7t}dt=\frac{1}{63}\left(-\frac{1}{2}{e}^{-6t}-\frac{4}{7}{e}^{-7t}\right)$$Note that we didn’t include the constants of integration in each of these because including them would just have introduced a term that would get absorbed into the complementary solution just as we saw when we were dealing with 2^nd order differential equations.
Finally, a particular solution for this differential equation is then,
$$\begin{array}{cc}{Y}_P& =u_1{y}_1+u_2{y}_2+u_3{y}_3\\ & =\frac{1}{18}\left(e^{3t}+2e^{2t}\right)e^{-3t}-\frac{1}{14}\left(3e^t+4t\right)e^{-t}+\frac{1}{63}\left(-\frac{1}{2}{e}^{-6t}-\frac{4}{7}{e}^{-7t}\right)e^{6t}\\ & =-\frac{1}{6}+\frac{5}{49}{e}^{-t}-\frac{2}{7}t{e}^{-t}\end{array}$$The general solution is then,
$$y\left(t\right)=c_1{e}^{-3t}+c_2{e}^{-t}+c_3{e}^{6t}-\frac{1}{6}+\frac{5}{49}{e}^{-t}-\frac{2}{7}t{e}^{-t}$$

We’re only going to do a single example in this section to illustrate the process more than anything so with that we’ll close out this section.