Question: When we use the cylindrical coordinate system $(r, \theta, z)$ where $r$ is the distance from the point in the $xy$-plane, $\theta$ is the angle with the $x$ axis and $z$ is the heigth. As can been seen in the picture

[![enter image description here](https://i.stack.imgur.com/lKIQb.png)](https://i.stack.imgur.com/lKIQb.png)

I hav a vector field described by $(0,U\_{\theta}(r),U\_z)$ but how can the angle differ when $r$ is always zero?
When $r$ is zero, then there is no difference between different angles. Or am i wrong?

Some help would be great.

Answer: 
**Hint:**

You discover the rules by working backwards.

First consider polynomials, say of the third degree, like $ax^3+bx^2+cx+d$. The first derivative is a second degree polynomial, so that the LHS of a first order equation would yield

$$\alpha y+\beta y'=\alpha ax^3+(\alpha b+3\beta a)x^2+(\alpha c+2\beta b)x+(\alpha d+\beta c),$$ another cubic polynomial. Similarly,

$$\alpha y+\beta y'+\gamma y''=\alpha ax^3+(\alpha b+3\beta a)x^2+(\alpha c+2\beta b+6\gamma c)x+(\alpha d+\beta c+2\gamma b)$$ also third degree.

Now it is clear that for a linear equation of any degree, if the RHS is a polynomial of degree $d$, there will be another polynomial of degree $d$ which is a solution.

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For an exponential function, $e^{ax}$, 

$$\alpha y+\beta y'+\gamma y''=(\alpha +\beta a+\gamma a^2)e^{ax}$$ so that an exponential RHS is obtained from an exponential solution.

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You can play the same game with RHS of the form $e^{ax}\sin bx$ and $e^{ax}\cos bx$.

If you are familiar with complex numbers, you can even take a shortcut by considering $e^{(a+ib)x}$, which combines the two functions in one.
$$y'''(x)-2y''(x)+y'(x)=x+2$$
It's easy to determine the exponent of the function
try to substitute $y=z+ax^n$
$$y'''+.............+y'=x+2$$
$$(z'''+....)+..............+(z'+nax^{n-1})=x+2$$
for y' we get the highest monomial degree in x so it must be equal to the degre of x on right side
$$x^{n-1}=x \implies n=2$$
So you have to try a polynomial of degree two
$$ax^2+bx+c$$
You observe that you have only derivatives on the left side so $c=0$

And $ax^2+bx$ is the right choice...