# Singular Points

• Jan 21st 2010, 03:50 AM
kaboose786
Singular Points
Need help with solving this question i have no idea how to even start so any help would be appreciated thanks

http://i47.tinypic.com/2rw3w1y.jpg
• Jan 21st 2010, 04:51 AM
shawsend
Why not just start by solving the DE:

$\displaystyle y'\pm\frac{2}{x}y=3$

via integrating factor and obtain the two solutions. What can you say about either solution at x=0?
• Jan 21st 2010, 11:20 AM
kaboose786
Quote:

Originally Posted by shawsend
Why not just start by solving the DE:

$\displaystyle y'\pm\frac{2}{x}y=3$

via integrating factor and obtain the two solutions. What can you say about either solution at x=0?

y do i need to solve that DE ? thats just the example in question i think i need to solve the DE in part a) which i dont knw how to
• Jan 22nd 2010, 04:34 AM
shawsend
Ok, then here's how I'd do (A):

$\displaystyle xy'+2y=3x,\quad y(0)=y_0$ and dividing by $\displaystyle x$:

$\displaystyle y'+2/x y=3$ for which the integrating factor is $\displaystyle x^2$ so we have:

$\displaystyle x^2(y'+2/x y)=3x^2$ which the left side is the differential of $\displaystyle x^2y$ so:

$\displaystyle \int d(x^2y)=\int 3x^2$ or:

$\displaystyle y(x)=x+\frac{c}{x^2}$ and if $\displaystyle y_0=0$ then I can let $\displaystyle c=0$ and obtain the solution $\displaystyle y=x$. However, if $\displaystyle y_0\neq 0$ say $\displaystyle 3$ for example, then I'd obtain $\displaystyle 3=0+\frac{c}{0}$ which has no solution.

Ok, so I'm a little uneasy about dividing through by $\displaystyle x$ in the first step since this is valid if $\displaystyle x\ne 0$ then obtain the solution and solve it for $\displaystyle x=0$.