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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?

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    $\begingroup$ linear equations must involve $y, y', y''$ etc. with coefficients that are (at worst) functions of $x$. terms like $yy'$ or $y^2$ are ruled out $\endgroup$ – citedcorpse Jun 8 '13 at 10:10
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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 $g(y_1+y_2)\neq g(y_1) + g(y_2)$).

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  • $\begingroup$ does this mean that linear differential equation has one y, and non-linear has two y, y'? $\endgroup$ – maycca Jun 21 '17 at 8:28
  • $\begingroup$ @Daniel Robert-Nicoud does the same thing apply for linear PDE? Wikipedia says PDE is linear if it is linear in dependent variable and its derivatives. $\endgroup$ – ramanujan Nov 20 '18 at 20:20
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    $\begingroup$ @ramanujan Yes, it does. $\endgroup$ – Daniel Robert-Nicoud Nov 20 '18 at 21:55
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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:

  1. 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.

  2. 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. :-)

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    $\begingroup$ I have edited your answer for better readability. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Jesse P Francis Nov 13 '15 at 4:04
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One could define a linear differential equation as one in which linear combinations of its solutions are also solutions.

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    $\begingroup$ your statement is true for only homogeneous LDE? For nonhomogeneous it is false. And I think it is also false in PDE. $\endgroup$ – ramanujan Nov 20 '18 at 20:16
  • $\begingroup$ @ramanujan What is an example of a non-homogeneous LDE or PDE whose solutions' linear combinations are also solutions? $\endgroup$ – Geremia Nov 20 '18 at 22:29
  • $\begingroup$ (do you mean 'solution of linear combinations are not solutions?') Simplest example is $y^{\prime}= 2$ and take both solutions $y=2 x$. $\endgroup$ – ramanujan Nov 21 '18 at 4:21
  • $\begingroup$ @ramanujan $y'=2$ is non-homogeneous? $\endgroup$ – Geremia Nov 21 '18 at 4:37
  • $\begingroup$ Cited sources says first order LDE is nonhomogeneous if $y^{\prime} + p(x)y = q(x)$, if $q(x)$ is not identically zero. here and here $\endgroup$ – ramanujan Nov 21 '18 at 7:38
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Linear Differential equations are those in which the dependent variable and its derivatives appear only in first degree and not multiplied together

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    $\begingroup$ $y'''+y''+y=e^x$ is linear ;) The degree is irrelevant. What's important is "not multiplied together" $\endgroup$ – Scientifica Aug 28 '16 at 6:21
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Because highest order derivative is multiplied with dependent variable $y$. Like $y y'$.

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