Singular Points

**kaboose786** · January 21st 2010, 03:50 AM

Need help with solving this question i have no idea how to even start so any help would be appreciated thanks

**shawsend** · January 21st 2010, 04:51 AM

Why not just start by solving the DE:

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

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

**kaboose786** · January 21st 2010, 11:20 AM

Originally Posted by shawsend

Why not just start by solving the DE:

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

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

**shawsend** · January 22nd 2010, 04:34 AM

Ok, then here's how I'd do (A):

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

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

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

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

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

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

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