Linear Differential Equation Interpolation

Homogeneous Solutions

If the functions \(\{y_1,\ldots,y_n\}\) are linearly independent, then the \(n\)th-order linear differential equation

has the solution

The proof is relatively trivial. Substituting the solution into the determinant makes the first row a linear combination of the other rows. Therefore the determinant will be zero.

Particular Solutions

The goal of interpolating a particular solution \(y_p\) for a differential equation is that when we plug in \(y_p\) to the differential equation we get the forcing function. So… we just do that.

If the functions \(\{y_1,\ldots,y_n,\}\) are linearly independent, then the \(n\)th-order linear differential equation

has the solution

\(y_p\) need not be independent from the homogeneous solutions. If it is, the determinant of the right-hand side will be zero leading back to the other case.

Examples

1. Find a linear differential equation for which \(y_1=e^{at}\) and \(y_2=e^{bt}\) are homogeneous solutions. We know the answer will be \((D-a)(D-b)y=0\), but let’s try the determinant.

   \[\begin{equation}\begin{vmatrix} y&y'&y''\\ e^{at}&ae^{at}&a^2e^{at}\\ e^{bt}&be^{bt}&b^2e^{bt} \end{vmatrix}=0 \end{equation}\]

   Pulling the exponentials out,

   \[\begin{equation} e^{at}e^{bt}\begin{vmatrix} y&y'&y''\\ 1&a&a^2\\ 1&b&b^2 \end{vmatrix}=0 \end{equation}\]

   General exponentials are always nonzero, so we can safely divide them out.

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ 1&a&a^2\\ 1&b&b^2 \end{vmatrix}=0 \end{equation}\]

   Using determinant properties to simplify just a bit,

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ 1&a&a^2\\ 0&b-a&b^2-a^2 \end{vmatrix}=0 \end{equation}\] \[\begin{equation} (b-a)\begin{vmatrix} y&y'&y''\\ 1&0&-ab\\ 0&1&b+a \end{vmatrix}=0 \end{equation}\]

   Finally, expanding along the top row,

   \[\begin{equation} (b-a)\left(\begin{vmatrix}0&-ab\\1&b+a\end{vmatrix}y-\begin{vmatrix}1&-ab\\0&a+b\end{vmatrix}y'+\begin{vmatrix}1&0\\0&1\end{vmatrix}y''\right)=0 \end{equation}\] \[\begin{equation} (b-a)\left(y''-(a+b)y'+aby\right)=0 \end{equation}\]

   Assuming \(a\neq b\), we can divide out to get

   \[\begin{equation} y''-(a+b)y'+aby=0 \end{equation}\]

   As expected.

   You will get the same result if \(a=b\), and \(y_2=e^{at}(mt+c)\) or something of that sort.

2. Find a linear differential equation for which \(y_1=x^{r}\) and \(y_2=x^{r}\ln(x)\) are homogeneous solutions.

   Setting up the determinant,

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x^r&rx^{r-1}&r(r-1)x^{r-2}\\ x^r\ln(x)&rx^{r-1}\ln(x)+x^{r-1}&r(r-1)x^{r-2}\ln(x)+rx^{r-2}+(r-1)x^{r-2} \end{vmatrix}=0 \end{equation}\]

   The third row has the second row times \(\ln(x)\) added to it, so we can subtract it out.

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x^r&rx^{r-1}&r(r-1)x^{r-2}\\ 0&x^{r-1}&(2r-1)x^{r-2} \end{vmatrix}=0 \end{equation}\]

   Factoring out,

   \[\begin{equation} x^{r-2}x^{r-2}\begin{vmatrix} y&y'&y''\\ x^2&rx&r(r-1)\\ 0&x&(2r-1) \end{vmatrix}=0 \end{equation}\]

   We can just divide out the powers of \(x\), since \(x=0\) is already out of our domain by having a factor of \(\ln(x)\) as one of our solutions.

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x^2&0&-r^2\\ 0&x&(2r-1) \end{vmatrix}=0 \end{equation}\]

   When expanded, we get

   \[\begin{equation} x^3y''-(2r-1)x^2y'+r^2xy=0 \end{equation}\]

   Dividing out an \(x\),

   \[\begin{equation} x^2y''-(2r-1)xy'+r^2y=0 \end{equation}\]

3. Let’s now find a differential equation for which the general solution is \(y=c_1\cos(x)+c_2\sin(x)+x\sin(x)+\cos(2x)\). This would give us \(y_1=\cos(x)\), \(y_2=\sin(x)\), and \(y_p=x\sin(x)+\cos(2x)\).

   We start with zero on the right side even though we want a particular solution, since these simplification steps would be identical on both sides.

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ \cos(x)&-\sin(x)&-\cos(x)\\ \sin(x)&\cos(x)&-\sin(x)\\ \end{vmatrix}=0 \end{equation}\] \[\begin{equation}\begin{vmatrix} y&y'&y''\\ \cos^2(x)&-\sin(x)\cos(x)&-\cos^2(x)\\ \sin^2(x)&\sin(x)\cos(x)&-\sin^2(x)\\ \end{vmatrix}=0 \end{equation}\] \[\begin{equation}\begin{vmatrix} y&y'&y''\\ \cos^2(x)&-\sin(x)\cos(x)&-\cos^2(x)\\ \sin^2(x)+\cos^2(x)&0&-(\sin^2(x)+\cos^2(x))\\ \end{vmatrix}=0 \end{equation}\] \[\begin{equation}\begin{vmatrix} y&y'&y''\\ 0&-\sin(x)\cos(x)&0\\ 1&0&-1\\ \end{vmatrix}=0 \end{equation}\] \[\begin{equation}\begin{vmatrix} y&y'&y''\\ 0&1&0\\ -1&0&1\\ \end{vmatrix}=0 \end{equation}\]

   Expanding along the second row,

   \[\begin{equation}\begin{vmatrix} y&y''\\ -1&1\\ \end{vmatrix}=0 \end{equation}\]

   It is now a good time to get the particular solution determinants. By linearity, we can deal with them separately.

   \[\begin{equation}\begin{vmatrix} y&y''\\ -1&1\\ \end{vmatrix}=\begin{vmatrix} x\sin(x)&-x\sin(x)+2\cos(x)\\ -1&1\\ \end{vmatrix}+\begin{vmatrix} \cos(2x)&-4\cos(2x)\\ -1&1\\ \end{vmatrix} \end{equation}\] \[\begin{equation} y''+y=2\cos(x)-3\cos(2x) \end{equation}\]

4. And as a final example, \(y_1=x+1\), and \(y_2=e^x\).

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x+1&1&0\\ e^x&e^x&e^x \end{vmatrix}=0 \end{equation}\]

   Dividing out \(e^x\),

   \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x+1&1&0\\ 1&1&1 \end{vmatrix}=0 \end{equation}\] \[\begin{equation} \begin{vmatrix} y&y'&y''\\ x+1&1&0\\ -x&0&1 \end{vmatrix}=0 \end{equation}\] \[\begin{equation} \begin{vmatrix} 1&0\\0&1 \end{vmatrix}-y'\begin{vmatrix} x+1&0\\-x&1 \end{vmatrix}+y''\begin{vmatrix} x+1&1\\ -x&0 \end{vmatrix}=0 \end{equation}\] \[\begin{equation} xy''-(x+1)y'+y=0 \end{equation}\]

Note that this is not limited to second-order equations. Evaluating the determinant just gets factorially more tedious as you add more solutions and I didn’t want to deal with that because I am incredibly lazy.