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Mathematics 7 Online
OpenStudy (anonymous):

I need help identifying what type of differential equation theses examples are (simple,separable,linear, or homogeneous) dy/dx=3y+x^2 dy/dx=3yx^2 dy/dx=3x+x^2 dy/dx=3y+1 dy/dx=(3y+x)/(y-x) dy/dx=(3y+x^2)/(y-x) dy/dx=(3y+1)/(1-x)

OpenStudy (jamesj):

Rather than give you the answers, much better to ask you: what are the definitions and examples of each of those terms. Then I'd be happy to analyze a couple of these equations with you to see whether or not they meet those defintions

OpenStudy (anonymous):

okay great this is what i have so far dy/dx=3yx^2 is a separable differential equation dy/dx=3y+x^2 is linear d.e. dy/dx=3x+x^2 is a simple d.e. dy/dx=(3y+1)/(1-x) is a separable and homogenous d.e

OpenStudy (jamesj):

All of those are correct.

OpenStudy (anonymous):

alright awesome so the ones i am unsure on are dy/dx=3y+1 maybe a simple d.e? dy/dx=(3y+x)/(y-x) dy/dx=(3y+x^2)/(y-x) our teacher told us some of theses may be ones we haven't discussed

OpenStudy (jamesj):

Of these three: the first is linear, separable and inhomogeneous; the second and third are not simple, non-linear, inhomogeneous, and not separable.

OpenStudy (jamesj):

The last one is separable.

OpenStudy (jamesj):

That is , the last one in your original post.

OpenStudy (anonymous):

how do you know if a d.e. is linear i think im alil confused on that

OpenStudy (jamesj):

An equation is linear if all the powers of y, y', y'', y''', ... are only one. So y' + y = x is linear But y' + y^2 = 0 is not, nor y' + 1/y = x You have to be careful though and write things out (on paper or in your mind), because this equation is linear, for instance: y'/y + x = x^2

OpenStudy (anonymous):

oh okay that helps clear some things up for me, thanks!

OpenStudy (jamesj):

Technically, get all the terms of y one side of the equals sign and set them equal to zero. If you can substitute y = ky where k is a scalar into the equation and keep equality, then the equation is linear. For instance in my first equation above (ky)' + ky = 0 => k(y' + y) = 0 => y' + y = 0 so the equation is linear. But for y' + y^2 = 0, (ky)' + (ky)^2 = ky' + k^2.y^2 = 0 from which we can't deduce the original form

OpenStudy (anonymous):

Awesome! that helps alot i didn't take good notes on linear d.e. thanks again for your help

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