Linear vs nonlinear differential equation
How to distinguish linear differential equations from nonlinear ones? I know, that e.g.: $$ y''-2y = \ln(x) $$ is linear, but $$ 3+ yy'= x - y $$ is nonlinear. Why?
Solution 1:
Linear differential equations are those which can be reduced to the form $Ly = f$, where $L$ is some linear operator.
Your first case is indeed linear, since it can be written as:
$$\left(\frac{d^2}{dx^2} - 2\right)y = \ln(x)$$
While the second one is not. To see this first we regroup all $y$ to one side:
$$y(y'+1) = x - 3$$
then we simply notice that the operator $y\mapsto g(y) = y(y'+1)$ is not linear (for example we can take two functions $y_1$ and $y_2$ and notice that $g(y_1+y_2)\neq g(y_1) + g(y_2)$).
Solution 2:
If the equation would have had $\ln (y)$ on the right, that also would have made it non-linear, since natural logs are non-linear functions. Remember that this has its roots in linear algebra: $y=mx+b$. You can analyse functions term-by-term to determine if they are linear, if that helps. The first time a term is non-linear, then the entire equation is non-linear.
Remember that the $x$s can pretty much do or appear however they want, since they're independent. Which means if you can't tell just by glancing, try to group all your $y$ terms to one side and then analyse them. Makes it much easier.
See, I was also overthinking this, but realised you have to go back to those definitions we're given.
Two criteria for linearity:
The dependent variable y and its derivatives are of first degree; the power of each y is 1. $\frac{dy}{dx}$; yes. $(\frac{dy}{dx})^4$, no.
Each coefficient depends only on the independent variable $x$.
$yy'$ makes it nonlinear as has been said, because that coefficient on $y'$ is not in $x$. Had that coefficient been a constant, you would have been correct to call it linear, since constants can be functions of $x$. Like, $f(3)=x$. Its graph is a line, i.e. linear function.
Always go back to the definitions. :-)