Understanding Engineering Mathematics

Solving these two equations givesA=1,B=−1 and the particular solution satisfying
the initial conditions is

y=e−x−e−^2 x

Problem 15.15

Find the general solution of the DE

y′′− 6 y′Y 9 y= 0

Substitutingy=eλxgives, in this case, the AE

λ^2 − 6 λ+ 9 = 0

or
(λ− 3 )^2 = 0

We therefore have two equal roots,λ=3 (rather than a single root ofλ=3!). One solution
of the DE is therefore obviously

y=e^3 x

But we expect two. On such occasions we try to guess other, similar, types of solution.
While there is a full justification for our guesswork in the advanced theory of differential
equations here I ask you to simply accept and confirm for yourself that the simplest guess
that works is

y=xe^3 x

This gives

y′=( 3 x+ 1 )e^3 x

y′′= 3 e^3 x+ 3 ( 3 x+ 1 )e^3 x

=( 9 x+ 6 )e^3 x

Substituting in the equation gives

( 9 x+ 6 )e^3 x− 6 ( 3 x+ 1 )e^3 x+ 9 xe^3 x≡ 0

so this is indeed a solution. The general solution in this case is therefore

y=Axe^3 x+Be^3 x

=(Ax+B)e^3 x

Problem 15.16

Find the general solution of the equation

y′′Y 2 y′Y 2 y= 0