1 problem found
Show that the second-order differential equation \[ x^2y''+(1-2p) x\, y' + (p^2-q^2) \, y= \f(x) \,, \] where \(p\) and \(q\) are constants, can be written in the form \[ x^a \big(x^b (x^cy)'\big)' = \f(x) \,, \tag{\(*\)} \] where \(a\), \(b\) and \(c\) are constants.
Solution: Consider $x^a \big(x^b (x^cy)'\big)'$ then \begin{align*} x^a \big(x^b (x^cy)'\big)' &= x^a \big (bx^{b-1}(x^c y)'+x^b(x^cy)'' \big ) \\ &= x^a \big (bx^{b-1} (cx^{c-1}y + x^c y') + x^b(c(c-1)x^{c-2}y + 2cx^{c-1}y' + x^cy'') \\ &= x^{a+b+c}y'' + (2cx^{c-1+b+a}+bx^{c+b-1+a})y'+(c(b+c-1))x^{a+b+c-2} y \end{align*} So we need: \begin{align*} &&& \begin{cases} a+b+c &= 2 \\ 2c+b &= 1-2p \\ c(b+c-1) &= p^2-q^2 \end{cases} \\ \Rightarrow && c((1-2p)-2c+c-1) &=p^2-q^2 \\ \Rightarrow && c^2+2pc &= q^2-p^2 \end{align*}