# linear and non-linear equations

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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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Attachment 886898

Last edited by ddsizebra; 1 year ago

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**mqb2766**)Attachment not found

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#4

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you sure? I see it on my account

**ddsizebra**)you sure? I see it on my account

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#5

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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#6

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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.

**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.

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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

**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

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#8

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sorry still don't understand this

**ddsizebra**)sorry still don't understand this

http://mathonline.wikidot.com/the-pr...-superposition

Last edited by mqb2766; 1 year ago

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#9

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

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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(Original post by

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

**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

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#11

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

**ddsizebra**)I understand what you've explained but cannot relate this to the attached image in the question

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#13

(Original post by

Assume you had a solution

**ddsizebra**)Assume you had a solution

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

Last edited by mqb2766; 1 year ago

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