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Line 125: − =Non-homogeneous and Non-Linear Equations= − − The theory of non-homogenous equations is incredibly difficult and intricate; the theory of non-linear equations is practically a wilderness of weak theorems dealing with special cases. Here, we outline a few techniques which sometimes prove useful. − − ==Infinite Series== − − Keeping in mind how many functions can be represented with infinite sums created with Taylor series, we assume a function has an infinite series representation. − − For example, let's solve <math>y''-xy'+x^2y=e^x</math> by assuming there exists <math>a_n</math> such that <math>y=\sum_{n=0}^\infty a_nx^n</math>. Then our differential equation becomes − − <math>(2a_2+(3)(2)a_3x+(4)(3)a_4x^2+\dots)-x(a_1+2a_2x+3a_3x^2+\dots)+x^2(a_0+a_1x+a_2x^2+\dots) = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots</math> − − Grouping like terms, we have − − <math>(2a_2)+(6a_3-a_1)x+(12a_4-2a_2+a_0)x^2 + \dots </math> − − so that for <math>n\geq 2</math>, the coefficient of <math>x^n</math> on the right hand side is <math>(n+1)(n+2)a_{n+2}-na_n+a_{n-2}</math>. So we need only set this equal to <math>1/(n!)</math> to find an infinite series representation of our solution. If we're given the values of <math>y(0), y'(0) \ </math>, as we frequently are in applications (remember what we said about having two "degrees of freedom" in a DEQ involving second-order derivatives?) we can solve this completely. Let's say <math>y(0)=0, y'(0)=1</math>. We have − − <math>a_0=0 \ </math> − <math>a_1=1 \ </math> − <math>a_2 = 1/2 \ </math>, since <math>2a_2=1 \ </math> − <math>a_3 = 1/3 \ </math>, since <math>6a_3-a_1=1 \ </math> − and so on. − − This method may not seem particularly useful. After all, unless we can find a closed form solution for the <math>a_n</math>, we are going to have to keeping solving this linear equations in <math>a_n</math> forever to get the solution. However, it may happen, on occasion, that we ''can'' find such a closed form representation. Or, it may happen that we find no closed form, but recognize it as similar or equivalent to a function which has been studied for decades or centuries, so that all the properties we wish to know about are easily available to us. Or, we may not ''care'' about a closed form - near 0, those higher powers will become very small, and a low-order approximation becomes highly accurate. − − ==Functional Transformations== − − It would be so convenient if we had some kind of easily-performed transformation which took one function, ''f'', and gave us another function, <math>f^*</math>, which had the property that the transform of <math>f'</math> was not a derivative of ''f'' but some algebraic multiple or something like that of ''f''. Then we could transform complicated differential equations into algebraic equations. − − It seems a tall order, but fortunately, there are functional transforms which behave in just this way. − − ===The Fourier Transform=== − − ===The Laplace Transform===
removed a section - material is too advanced for MV calc course
It should come as no surprise by now to learn that given an equation <math>y^{(n)}+c_{n-1}y^{(n-1}+\dots+c_0=0</math>, a general solution is given by <math>y=a_1e^{r_1x}+a_2e^{r_2x}+\dots+a_ne^{r_nx} \ </math>, where the <math>r_j \ </math> are roots of the polynomial <math>\sum_{k=0}^n c_kt^k \ </math>. If two roots <math>r_i=r_j \ </math> are equal, then <math>e^{r_jx}, xe^{r_jx} \ </math> are both solutions; if three roots <math>r_i=r_j=r_k \ </math> are equal, then <math>e^{r_kx}, xe^{r_kx}, x^2e^{r_kx} \ </math> are all solutions; if four roots <math>r_i=r_j=r_k=r_l \ </math> are equal, then <math>e^{r_lx}, xe^{r_lx}, x^2e^{r_lx}, x^3e^{r_lx} \ </math> are all solutions, and so on.
It should come as no surprise by now to learn that given an equation <math>y^{(n)}+c_{n-1}y^{(n-1}+\dots+c_0=0</math>, a general solution is given by <math>y=a_1e^{r_1x}+a_2e^{r_2x}+\dots+a_ne^{r_nx} \ </math>, where the <math>r_j \ </math> are roots of the polynomial <math>\sum_{k=0}^n c_kt^k \ </math>. If two roots <math>r_i=r_j \ </math> are equal, then <math>e^{r_jx}, xe^{r_jx} \ </math> are both solutions; if three roots <math>r_i=r_j=r_k \ </math> are equal, then <math>e^{r_kx}, xe^{r_kx}, x^2e^{r_kx} \ </math> are all solutions; if four roots <math>r_i=r_j=r_k=r_l \ </math> are equal, then <math>e^{r_lx}, xe^{r_lx}, x^2e^{r_lx}, x^3e^{r_lx} \ </math> are all solutions, and so on.