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linear and non-linear equations

1a is said to be linear and 1b is said to be non-linear but I can't see the difference or as to why they've been described so... can someone clarify please?

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(edited 4 years ago)
Reply 1
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Reply 2
Original post by mqb2766
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you sure? I see it on my account
Reply 3
Original post by ddsizebra
you sure? I see it on my account

I can't see it. But an equation is linear if it satisfies the addition and scaling properties, if you've covered those. Otherwise it's non linear.
Reply 4
Non-linear in terms of DEs is when you have a y or dy/dx times a dy/dx or a d^2y/dx^2 or whatever. It could also be if you had a (dy/dx)^2 or anything like that too.
Reply 5
Original post by Acman
Non-linear in terms of DEs is when you have a y or dy/dx times a dy/dx or a d^2y/dx^2 or whatever. It could also be if you had a (dy/dx)^2 or anything like that too.


the reasoning behind this is if you said y = az + bw, the linear differential equation would behave linearly - f(ax + by) = af(x) + bf(y) - whereas in a nonlinear you can see this won't happen
Reply 6
Original post by Acman
the reasoning behind this is if you said y = az + bw, the linear differential equation would behave linearly - f(ax + by) = af(x) + bf(y) - whereas in a nonlinear you can see this won't happen


sorry still don't understand this
Reply 7
Original post by ddsizebra
sorry still don't understand this

What are your two examples? For superposition (scaling and addition) see
http://mathonline.wikidot.com/the-principle-of-superposition
(edited 4 years ago)
Reply 8
lets consider something simpler dy/dx = 0 and y * dy/dx = 0
let's define two functions f(y) = dy/dx, g(y) = y * dy/dx
if something haves linearly it should follow the rule I stated above, so let's test it
let y = az + bw where z and w are new variables and a and b are constants
f(y) = f(az + bw) = d(az + bw)/dx, now since the differential operator is a linear operator we can rewrite this
= a * dz/dx + b * dw/dx, now let's compare this with out original function
= a * f(z) + b * f(w)

so we have f(az + bw) = a * f(z) + b * f(w) so f is linear, and thus dy/dx = 0 is linear

now if you carry out this process with g(y) = g(az + bw) you can see why it's not linear

a slightly simpler way to think of it is that something linear is of the form ay + b where a and b are constant, from this you can see that y * dy/dx is not linear because dy/dx is not necessarily a constant, but this is not a good way to think about it really because you can have dy/dx + xy which is still linear but doesn't follow the ay + b form I described
Reply 9
Original post by Acman
lets consider something simpler dy/dx = 0 and y * dy/dx = 0
let's define two functions f(y) = dy/dx, g(y) = y * dy/dx
if something haves linearly it should follow the rule I stated above, so let's test it
let y = az + bw where z and w are new variables and a and b are constants
f(y) = f(az + bw) = d(az + bw)/dx, now since the differential operator is a linear operator we can rewrite this
= a * dz/dx + b * dw/dx, now let's compare this with out original function
= a * f(z) + b * f(w)

so we have f(az + bw) = a * f(z) + b * f(w) so f is linear, and thus dy/dx = 0 is linear

now if you carry out this process with g(y) = g(az + bw) you can see why it's not linear

a slightly simpler way to think of it is that something linear is of the form ay + b where a and b are constant, from this you can see that y * dy/dx is not linear because dy/dx is not necessarily a constant, but this is not a good way to think about it really because you can have dy/dx + xy which is still linear but doesn't follow the ay + b form I described


I understand what you've explained but cannot relate this to the attached image in the question
Original post by ddsizebra
I understand what you've explained but cannot relate this to the attached image in the question

Can you upload again to your reply?
Reply 11
Original post by mqb2766
Can you upload again to your reply?


20200316_063901-compressed.jpg.jpeg
Original post by ddsizebra
20200316_063901-compressed.jpg.jpegAssume you had a solution

Have you tried to apply superposition as per the other link or
http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx
Note the odes are nonhomogeneous.
https://sites.oxy.edu/ron/math/341/10/ws/08.pdf
http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx
(edited 4 years ago)

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