Find a general solution? check answer using substitution.

Question

(a)y" + 4piy' + 4pi^2y=0
(b)y" + 2.6y' +1.69y = 0

1812 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-25T10:01:05

(a) #y=ae^(-2pix)# and (2) #y=ae^(-1.3x)#

Let the solution to #y''+4piy'+4pi^2y=0# be #y=e^(kx)#.

Then #y'=ke^(kx)# and #y''=k^2e^(kx)#

and #y"+4piy'+4pi^2y=0# becomes

#k^2e^(kx)+4pike^(kx)+4pi^2e^(kx)=0#

or #(k^2+4pik+4pi^2)e^(kx)=0#

or #(k+2pi)^2=0#

i.e. #k=-2pi#

and solution is #y=e^(-2pix)#. Infact, #y=ae^(-2pix)# too will be a solution, where #a# is a constant.

Therefore solution of #y"+2.6y'+1.69y=0# is #y=ae^(-1.3x)#

then #y'=-1.3ae^(-1.3x)# and #y''=1.3xx1.3xxae^(-1.3x)=1.69ae^(-1.3x)#

substituting #1.69ae^(-1.3x)-2.6xx1.3ae^(-1.3x)+1.69ae^(-1.3x)#

= #1.69ae^(-1.3x)-3.38ae^(-1.3x)+1.69ae^(-1.3x)=0#

Note: If you get a quadratic equation of type #(k-alpha)(k-beta)=0# in place of #(k+c)^2=0#, the solution would be #a_1e^(alphax)+a_2e^(betax)#.