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Show that if we have two independent solutions to a 3rd order homogenous linear differential equation y'''+a_{2}y''+a_{1}y'+a_{0}=0 we can always find a third solution which is independent from the two given solutions.

Here is my proof:

since y_{2}and y_{1}are linearly independent, there exists a non-constant function u such that y_{2}=uy_{1}.

Now, If we plug y_{2}into the equation, we can reduce the order of the original equation to a 2nd order linear ODE of the form a(x)v+b(x)v'+c(x)'' = 0 where:

[tex]u=y_2/y_1=\int{v(x)dx}[/tex]

Now, If I set v=wy_{3}, and I suppose t=w', then doing the same process I can reduce the ODE a(x)v+b(x)v'+c(x)v''=0 into a first order linear ODE of the form r(x)t+s(x)t'=0.

This ODE is always solvable and the solution to this ODE is:

[tex]t=Ce^{\int{-r(x)/s(x)dx}}[/tex]

so, w would be:

[tex]w=C\int{e^{\int{-r(x)/s(x)dx}}}[/tex]

Having w and v, we can find y_{3}=v/w.

This shows us that y_{3}can be found, but I don't know how to show that y_{3}is necessarily independent from y_{1}and y_{2}.

Any idea that completes my proof would be highly appreciated.

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# Does my proof make sense?

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