Doubt related to formation of a differential equation

[cheapstrike]

Homework Statement

Find the order of the differential equation of y=C₁sin²x+C₂cos2x+C₃.

Homework Equations

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The Attempt at a Solution

[/B]
I read in my book that the order of the differential equation is equal to the number of arbitrary constants but the answer given is 2.

Btw I have uploaded two pics with two methods I tried.
1st Pic - http://imgur.com/Py7DTgp
2nd Pic - http://imgur.com/5pdcWq1

In first picture, I calculated upto 3rd differential and obtained a differential equation.

In second picture, I differentiated both sides w.r.t. x and then sent the sin2x term, which I was getting in RHS, to LHS and wrote (1/sin2x) as cosec2x. Then I differentiated both sides again w.r.t. x. In this way, both C1 and C2 which remained after calculating 1st derivative become zero.

Which one is correct method? If it's neither, then what's the right method?

 

[Orodruin]

I read in my book that the order of the differential equation is equal to the number of arbitrary constants but the answer given is 2.

Your function only has two arbitrary constants. The terms are not linearly independent.

 

[cheapstrike]

Your function only has two arbitrary constants. The terms are not linearly independent.

Can you elaborate please? I mean how is C₃ not an arbitrary constant?

 

[Orodruin]

Can you elaborate please? I mean how is C₃ not an arbitrary constant?

The functions that they multiply are linearly dependent. You can rewrite $\sin^2 x$ in terms of cos(2x) and a constant.

 

[cheapstrike]

The functions that they multiply are linearly dependent. You can rewrite $\sin^2 x$ in terms of cos(2x) and a constant.

Thanks

 

[Ray Vickson]

Homework Statement

Find the order of the differential equation of y=C₁sin²x+C₂cos2x+C₃.

Can you see how to re-write $C_1 \sin^2(x) + C_2 \cos 2x + C_3$ as $A \cos2x + B$? Working with the latter form is easier.

Note added in edit: I see that once again PF has fooled me, by only displaying post #4 after I pressed submitted my answer.