ddsizebra
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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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mqb2766
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ddsizebra
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(Original post by mqb2766)
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you sure? I see it on my account
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mqb2766
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(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.
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Acman
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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.
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Acman
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(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
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ddsizebra
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(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
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mqb2766
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(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-pr...-superposition
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Acman
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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
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ddsizebra
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(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
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mqb2766
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(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?
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ddsizebra
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(Original post by mqb2766)
Can you upload again to your reply?
Name:  20200316_063901-compressed.jpg.jpeg
Views: 17
Size:  17.8 KB
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mqb2766
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(Original post by ddsizebra)
Name:  20200316_063901-compressed.jpg.jpeg
Views: 17
Size:  17.8 KBAssume you had a solution
Have you tried to apply superposition as per the other link or
http://tutorial.math.lamar.edu/Class...rConcepts.aspx
Note the odes are nonhomogeneous.
https://sites.oxy.edu/ron/math/341/10/ws/08.pdf
http://tutorial.math.lamar.edu/Class...geneousDE.aspx
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