Exponential approach

In mathematics, the exponential approach is understood to be an approach to solving a linear differential equation with constant coefficients, the inhomogeneity of which has an exponential structure. The idea is that a particulate solution of a similar shape to the inhomogeneity then also exists. Such a solution approach reduces the differential equation to a linear system of equations . The idea for this approach goes back to Leonhard Euler .

formulation

A linear differential equation is given

${\displaystyle y^{(n)}(x)+\sum _{k=0}^{n-1}c_{k}y^{(k)}(x)=b(x)}$

with constant coefficients , wherein the inhomogeneity is the structure ${\displaystyle c_{0},\ldots ,c_{n-1}\in \mathbb {C} }$

${\displaystyle b(x)=e^{(\alpha +i\beta )x}\sum _{k=0}^{l}a_{k}x^{k}\ ,\ \alpha ,\beta \in \mathbb {R} \ ,\ a_{0},\ldots ,a_{l}\in \mathbb {C} }$

owns. Also denote the zero order of with respect to the characteristic polynomial of the associated homogeneous equation ${\displaystyle j\in \mathbb {N} _{0}}$ ${\displaystyle \alpha +i\beta }$

${\displaystyle \chi (\lambda ):=\lambda ^{n}+\sum _{k=0}^{n-1}c_{k}\lambda ^{k}\ .}$

Then there is a special solution of the form ${\displaystyle y_{sp}}$

${\displaystyle y_{sp}(x)=e^{(\alpha +i\beta )x}\sum _{k=j}^{l+j}b_{k}x^{k}\ ,\ b_{j},\ldots ,b_{l+j}\in \mathbb {C} \ .}$

example

Consider the linear differential equation

${\displaystyle y''(x)+y(x)=xe^{ix}\ .}$

Now is the first order zero of the polynomial . So according to the above theorem there is a special solution of the shape ${\displaystyle i}$ ${\displaystyle \chi (\lambda )=\lambda ^{2}+1}$

${\displaystyle y_{sp}(x)=(ax+bx^{2})e^{ix}\ .}$

Out

${\displaystyle \ y_{sp}'(x)=(a+2bx)e^{ix}+i(ax+bx^{2})e^{ix}}$

and

${\displaystyle \ y_{sp}''(x)=2be^{ix}+2i(a+2bx)e^{ix}+i^{2}(ax+bx^{2})e^{ix}}$

is obtained from the differential equation

${\displaystyle (2b+2ai+4bix)e^{ix}=xe^{ix}\ .}$

Comparison of coefficients provides the determining equations

${\displaystyle 2b+2ai=0\ ,\ 4bi=1\ ,}$

which and implies. So is ${\displaystyle a={\tfrac {1}{4}}}$ ${\displaystyle b=-{\tfrac {1}{4}}i}$

${\displaystyle y_{sp}(x)=\left({\frac {1}{4}}x-{\frac {i}{4}}x^{2}\right)e^{ix}}$

a special solution to the above inhomogeneous differential equation.

literature

• Harro Heuser: Textbook of Analysis Part 1 . 5th edition. Teubner-Verlag 1988, ISBN 3-519-42221-2 , pp. 413-428.