If the equation looks like this: y = 2x +6 it is in what form?
2 answers:
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Given a solution 
, we can attempt to find a solution of the form 
. We have derivatives
Substituting into the ODE, we get
Setting
, we end up with the linear ODE
Multiplying both sides by
, we have
and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to
, we get
Now solve for
:
So you have
and given that
, the second term in 
is already taken into account in the solution set, which means that 
, i.e. any constant solution is in the solution set.
Substituting into the ODE, we get
Setting
Multiplying both sides by
and noting that
we can write the ODE as
Integrating both sides with respect to
Now solve for
So you have
and given that