If # y=ce^(2x)+De^(-2x) # then show that # y'' -4y = 0 #?

3 Answers
Aug 15, 2017

#y''-4y=0#

Explanation:

This is an inverse problem. Given a solution, find a candidate differential equation which has is as solution.

Assuming the differential equation is second-order linear homogeneous such as

#c_1 y'' +c_2 y'+c_3 y =0#

after substituting the solution we have

#(4 c_1 - 2 c_2 + c_3) D_0 e^(-2 x) + C_0 (4 c_1 + 2 c_2 + c_3) e^(2 x)=0#

This relationship must be true for all #x# so

#{(4 c_1 - 2 c_2 + c_3=0),(4 c_1 + 2 c_2 + c_3=0):}#

now solving for #c_1,c_2# we obtain

#c_1 = -c_3/4, c_2 = 0# so the differential equation is

#-c_3/4 y'' +c_3 y =0# or

#y''-4y=0#

Aug 15, 2017

Refer to the Explanation.

Explanation:

We note that, the given eqn. #y=Ce^(2x)+De^(-2x)......(1),# contains

2 arbitrary constants.

Therefore, the reqd. Diff. Eqn. must be of Second Order.

To find it, we diif. #(1)# twice.

# y=Ce^(2x)+De^(-2x) rArr dy/dx=(Ce^(2x))2+(De^(-2x))(-2), or,#

# dy/dx=2{Ce^(2x)-De^(-2x)},#

#:. (d^2y)/dx^2=d/dx[2{Ce^(2x)-De^(-2x)}],#

#=2[(Ce^(2x))(2)-(De^(-2x))(-2)],#

# rArr (d^2y)/dx^2=4[Ce^(2x)+De^(-2x)]=4y,#

# rArr (d^2y)/dx^2-4y=0,# is the Desired Diff. Eqn.

Aug 15, 2017

# y'' -4y = 0 #

Explanation:

We have:

# y=ce^(2x)+De^(-2x) # .... [A}

As others have indicated we are not solving a second order Differentiation Equation with constant coefficients, but rather forming one given the solution.

Recognizing the solution is that of a second order Differentiation Equation with constant coefficients we can instantly write down the appropriate DE.

The Auxiliary Equation that produced this solution would require two distinct real solution, #m=2# and #m=-2)#

Hence the associated Auxiliary Equation would be:

# (m-2)(m+2) = 0 => m^2-4 = 0 #

Hence the DE associated with this Auxiliary Equation is:

# y'' + 0y'-4y = 0 #
# y'' -4y = 0 #