Some more based on forms
Method of Variation of Parameters #
$$ \begin{align*} a_{0}(x)y’’(x)+a_{1}(x)y’(x)+a_{2}(x)y(x)=r(x)\\ \\ y_c(x)=A_{1}y_{1}(x)+A_{2}y_{2}(x)\\ y_p(x)=A_{1}(x)y_{1}(x)+A_{2}(x)y_2(x)\\ \\ A_1(x)=-\int\left(\frac{y_{2}(x)g(x)}{W(x)}\right)dx \\ and\\ \\ B(x)=\int \left(\frac{y_{1}(x)g(x)}{W(x)}\right)dx\\ g(x)=\frac{r(x)}{a_{0}} \end{align*} $$
W is the Linear Dependence of Functions
Cauchy Euler equation #
Notes are by Sir, but can also refer to: https://www.math.utah.edu/~gustafso/s2014/3150/slides/cauchy-euler-de.pdf
Form #
$$ (a_{0}x^{n}D^{n}+a_{1}x^{n-1}D^{n-1} +…+a_{n-1}xD + a_{n})y = r(x)\\ $$
Simplify and Solve #
We can convert the differential equation with variable coefficients to a differential equation with constant coefficients by doing a substitution $x=e^{t}$
$$ \begin{align*} Let\\ D = \frac{d}{dx}\\ \theta = \frac{d}{dt} \end{align*} $$
Relation between operators #
First $$ \begin{align*}\\ x=e^{t};t=\ln x\\ \dd{y}{x}=\dd{y}{t} \dd{t}{x}\\ x D y = \theta y\\ x D = \theta \end{align*} $$
Second $$ \begin{align*} \dd{^{2}}{x^{2}}y=\dd{}{x} (\dd{y}{x})\\ =\dd{}{x}(\frac{1}{x} \dd{y}{t})\\ \frac{-1}{x^{2}} \dd{y}{t} + \frac{1}{x} \dd{}{x} (\dd{y}{t})\\ \frac{-1}{x^{2}} \dd{y}{t} + \frac{1}{x} \dd{}{t}(\dd{y}{t}) \dd{t}{x}\\ \frac{-1}{x^{2}} \dd{y}{t} + \frac{1}{x} \dd{^{2}}{t^{2}} \frac{1}{x}\\ \\ x^{2}D^{2}y=(\theta ^{2}- \theta)y\\ x^{2}D^{2} = \theta(\theta -1) \end{align*} $$
Generalized $$ x^{n}D^{n} = \theta(\theta -1)…(\theta-n+1) $$
Just substitute these in the original to get constant coefficients.
Legendre’s Linear Equation #
Form #
$$ (a_{0}(ax+b)^{n}D^{n}+a_{1}(ax+b)^{n-1}D^{n-1} +…+a_{n-1}(ax+b)D + a_{n})y = r(x)\\ $$
Simplify and Solve #
Here too, we can convert the differential equation with variable coefficients to a differential equation with constant coefficients by doing a substitution $ax+b=e^{t}$
Relation between operators #
$$ (ax+b)^{n}D^{n}=a^{n}\theta (\theta -1)…(\theta - n +1) $$
Just substitute these in the original to get constant coefficients.