# 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.