#y''+9y=0#, #y_1=sin3x# by using reduction of order method how can I solve this equation?

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Answer 1 · 2017-12-16T23:04:57

See below.

#y''+9y=0# is a linear homogeneous differential equation with constant parameters. The general solution for an equation of this kind is

#y = e^(lambda x)#

substituting we get

#(lambda^2+9)e^(lambda x) = 0#

but #e^(lambda x) ne 0# for all #x in RR# so

#lambda^2+9=0 rArr lambda = pm 3i# so

#y = C_1e^(3ix) + C_2e^(-3ix)#

but using the de Moivre's identity

#e^(ix) = cos x +i sin x#

and assuming that the solution is real, we easily could establish analogously

#y = C_3 sin(3x) + C_4 cos(3x)#

Answer 2 · 2017-12-16T23:12:18

#y=c_1*cos3x+c_2*sin3x#

If #y_1=sin3x# i solution of this differential equation, I used #y=usin3x# and #y''=u''sin3x+6u'cos3x-9usin3x# transformation.

Hence,

#u''sin3x+6u'cos3x-9usin3x+9usin3x=0#

#u''sin3x+6u'cos3x=0#

It reduced to linear differential equation in terms of #u'#

Consequently,

#(u'')/(u')+(6cos3x)/sin3x=0#

After integrating both sides,

#Lnu'+2Ln(sin3x)=Ln(-3c_1)#

#Ln[u'*(sin3x)^2]=Ln(-3c_1)#

#u'*(sin3x)^2]=-3c_1#

#u'=(-3c_1)/(sin3x)^2#

After integrating both sides,

#u=c_1*cot3x+c_2#

Thus,

#y=usin3x#

=#sin3x*(c_1*cot3x+c_2)#

=#c_1*cos3x+c_2*sin3x#