Question #ed4b8

1 Answer
Jan 16, 2018

#(d^2y)/dx^2-9y=0#
#y=c_1e^(3x)+c_2e^(-3x)#

Explanation:

I assume that you want to solve #(d^2y)/dx^2-9y=0#.

Let us assume for now that #y=e^(rx)#, where #r# is a constant. Substitute this value into the equation:
#d^2/dx^2(e^(rx))-9e^(rx)=0#
#r^2e^(rx)-9e^(rx)=0#
#e^(rx)(r^2-9)=0#

Since #e^(rx)≠0# for all #x# and #r#, it must be the case that
#r^2-9=0#
#r=±3#

Thus we have two particular solutions #e^(3x)# and #e^(-3x)#. Recall that any linear combination of two linearly independent solutions to a second-order linear homogeneous differential equation forms the general solution. Thus, the general solution to the equation is #y=c_1e^(3x)+c_2e^(-3x)#, where #c_1,c_2# are arbitrary constants.