Is the ODE xy'' + siny = 0 Linear or Homogeneous?

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SUMMARY

The ordinary differential equation (ODE) xy'' + sin(y) = 0 is not linear due to the presence of the sine function, which introduces non-linearity. According to the definition of linearity, the coefficients of the unknown function y(x) and its derivatives must depend solely on x. The ODE fails this criterion as the operator L[y] = x y'' + sin(y) does not satisfy the linearity condition L[a y_1 + b y_2] = a L[y_1] + b L[y_2]. Therefore, the ODE is also classified as non-homogeneous.

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Homework Statement


Is the following ODE linear? If so, is it homogeneous?

xy'' + siny = 0, where y = y(x)


Homework Equations


Linear = coefficients of unknown function y(x) and its derivatives only depend on x, not the unknown

Homogeneous: can be written in form y'' + p(x)y' + q(x)y = 0


The Attempt at a Solution



I'm confused in that the only way to get rid of the sin from the y is to put an arcsin in front of the y'' term. Can one just 'ignore' the sin and say it is linear and also homogeneous by the definitions above?

Thanks for any help.
 
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Write [tex]L[y] = x y'' + \sin(y)[/tex]. [tex]L[/tex] is called a differential operator (a function of a function), and the given ODE is the same as finding the y so that
[tex]L[y] = 0[/tex]

An ODE is linear if its differential operator ([tex]L[/tex] above) is linear: in other words, for any two functions [tex]y_1,y_2[/tex] and two numbers [tex]a,b[/tex] we have
[tex]L[a y_1 + b y_2] = a L[y_1] + b L[y_2][/tex]
So the given ODE cannot be linear, since
[tex]L[a y_1 + b y_2] = axy_1'' + bxy_2'' + \sin(a y_1 + b y_2)[/tex]
and
[tex]aL[y_1] + bL[y_2] = axy_1'' + bxy_2'' + a\sin(y_1) + b\sin(y_2)[/tex]
which are not equal.

Notice that if [tex]L[y] = a_0(x) y(x) + a_1(x) y' + \ldots + a_n(x) y^{(n)}(x)[/tex] then
[tex]L[/tex] is linear by the above definition.
 

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