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

2 Answers
Dec 16, 2017

See below.

Explanation:

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)

Dec 16, 2017

y=c_1*cos3x+c_2*sin3x

Explanation:

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