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

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The ordinary differential equation (ODE) xy'' + sin(y) = 0 is determined to be nonlinear due to the presence of the sine function, which does not meet the criteria for linearity. A linear ODE requires that the coefficients of the unknown function and its derivatives depend only on the independent variable, which is not the case here. The attempt to apply the definition of linearity shows that the differential operator L[y] does not satisfy the linearity condition. Consequently, the ODE is also classified as non-homogeneous because it cannot be expressed in the standard form of a homogeneous equation. Therefore, the conclusion is that the ODE is neither linear nor 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 L[y] = x y'' + \sin(y). L is called a differential operator (a function of a function), and the given ODE is the same as finding the y so that
L[y] = 0

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

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