Home page for accesible maths 6 Chapter 6 contents

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

6.31 Particular integrals

It is possible to solve this equation systematically, similarly to the method for first-order linear equations. However, the quickest way to find the particular integral is by a form of trial and error. For each q⁢(x)q(x) as in the left-hand column below, we (usually) look for a PI of the form given in the right-hand column.

q⁢(x)PIan⁢xn+an-1⁢xn-1+…+a0bn⁢xn+bn-1⁢xn-1+…+b0eκ⁢xC⁢eκ⁢x (κ\kappa not a root)eκ⁢xC⁢x⁢eκ⁢x (κ\kappa a single root)eκ⁢xC⁢x2⁢eκ⁢x (κ\kappa a double root)H⁢cos⁡κ⁢x+K⁢sin⁡κ⁢xC⁢cos⁡κ⁢x+D⁢sin⁡κ⁢x (κ⁢i\kappa{\rm i} not a root)H⁢cos⁡κ⁢x+K⁢sin⁡κ⁢xx⁢(C⁢cos⁡κ⁢x+D⁢sin⁡κ⁢x) (κ⁢i\kappa{\rm i} a root)\begin{array}[]{ll}q(x)&{\rm PI}\\ a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}&b_{n}x^{n}+b_{n-1}x^{n-1}+\dots+b_{0}\\ e^{\kappa x}&Ce^{\kappa x}\quad{\hbox{($\kappa$ not a root)}}\\ e^{\kappa x}&Cxe^{\kappa x}\quad{\hbox{($\kappa$ a single root)}}\\ e^{\kappa x}&Cx^{2}e^{\kappa x}\quad{\hbox{($\kappa$ a double root)}}\\ H\cos\kappa x+K\sin\kappa x&C\cos\kappa x+D\sin\kappa x\quad{\hbox{($\kappa{% \rm i}$ not a root)}}\\ H\cos\kappa x+K\sin\kappa x&x(C\cos\kappa x+D\sin\kappa x)\quad{\hbox{($\kappa% {\rm i}$ a root)}}\end{array}

Here, when we say “κ\kappa a root”, we mean that it is a root of the auxiliary equation a⁢s2+b⁢s+c=0as^{2}+bs+c=0, so that eκ⁢xe^{\kappa x} is a solution of the homogeneous equation.