Hi maths people I have question for test I write next week.
There are differential equations and you say what property they have.
But my issue is I maybe don't understand all property right.
For order I count maximum number of derivative line. By this I mean for example $y''' + y''$ maximum line is $3$ so this is 3th order.
Linear you check if exponent of $y$ or $y$ with lines is equal to $1$. Example $(y'')^4$ not linear, $y^2$ not linear, but $y+y''$ is linear.
Homogeneous you check if equation is equal with zero and check if function have.. I call it disturbing function. If it have disturbing function you have no homogenetic. I don't can explain good sorry but here is example:
$y'''+2y'' = 0$ this is homo because equal to zero and no disturbing function.
$y'''-6xy' = 2-3e^x$ this is no homo because there is disturbung function $2$
But I have question, what if this is $y'''-6xy' = 3e^x$ (so without $2$) instead? I think is homogeneous because $x$ and $y$ belong to equation so there is no disturbing function. Is this right?
But what is constant coefficient? I think coefficient is the thing factorized by the variables. When it is number, it is constant coefficient.
But I don't know.. can you please give example?
Here is summary I make examples (can you say if this is right?):
$y''' +2y'' -5y'+3y+2=0$, 3th order, linear, constant coefficients, no homo
$2xy+x^2y'=0$, 1st order, linear, no constant coefficients because muliply by $x$, homo
Can you please say if all is good? My friend also not sure we learn
together and this is only thing we must understanded then ready for
test in school! Thank you very much for read all my question!!
Best Answer
Wow, when I first read your post as an american I thought you "writing" the test meant you were the teacher and I was scared for your class.
For order, your heuristic method is not terrible but it will mislead you in cases like this, $y'' + y' + y = y''$. This is a first order equation but there are terms with two "derivative lines" as you put it.
For linearity, you need to substitute a linear combination of solutions and show that is also a solution. What this is means is that everywhere you see $y$ you replace by $\alpha u + \beta v$ where you assume $\alpha, \beta$ are constants and $u$ and $v$ both satisfy the equation you are considering. Example: $y''+y=0$ is linear because $(\alpha u + \beta v)'' +(\alpha u + \beta v) = \alpha u'' + \beta v'' + \alpha u + \beta v= \alpha u'' + \alpha u + \beta v'' +\beta v= \alpha(u''+u)+\beta(v''+v)=\alpha\cdot0 + \beta\cdot 0=0$.
For homogeneity, you need to show that if $u$ is a solution then so is $cu$ for any constant $c$. For example, $y''+y=0$, we substitute $(cu)''+cu=cu''+cu=c(u''+u)=c\cdot0=0$.
The equation you ask about in your original post is NOT homogeneous, let's test the method described above. We substitute $cu$ into the equation to get, $(cu)''' -6x(cu)'=c(u'''-6xu)=c(3e^x)\neq 3e^x$ unless $c=1$. However, this condition must hold for ALL $c$, thus the equation is not homogeneous.