Question

Is the following differential equation linear and if so, is it homogenous or nonhomgeneous?

\[yy'=t^3+ysin(3t)\]

Now, I rewrote this as: \[y'-\frac{ t^3 }{ y }= \sin(3t)\]

Answer 1

the book says it is nonlinear because it can't be written in the for y' + p(t)y = g(t) ... so my question is, did I mess up rewriting this somehow, or is it simply nonlinear because its p(t)/y and not p(t)y?

Answer 2

It's not linear because it contains a function of the dependent variable \(y\). You have to regard \(\dfrac{1}{y}\) as a function of \(y\).

Answer 3

Ok cool, so my algebra was fine, but because its multiplied by 1/y and not y, its nonlinear

Answer 4

that makes sense, I didn't think of 1/y as a function of y. I was just treating it a little carelessly (any y term will do!)

Answer 5

Right. The distinction between linear and nonlinear can be a bit confusing sometimes.

Something like \(y'-y=\sin x\) might look nonlinear at first because of the sine term, but since it's a function of \(x\) and not \(y\), it's a totally valid liner equation.

Answer 6

Thanks for clearing that up!

Answer 7

You're welcome!

Answer 8

What about something like: \[2ty+e^yy' = \frac{ y }{ t^2 + 4 }\] .. If I rewrite this \[y' - y(\frac{ e ^{-y} }{ t^2+4 } -2te ^{-y}) = 0 \]

Answer 9

seems of the right form, unless the big term is "cheating" because it has y terms in it? @SithsAndGiggles

Answer 10

\(e^y\) is a red flag - another function of \(y\). You have to be able to completely isolate a \(y\) from that second term to be able to call the equation linear.

Answer 11

I was thinking that might be the case. Thanks again

Answer 12

No problem