Answer:

d2ydx2+1xdydx=12ln|x|x2  

x2d2ydx2+xdydx=12ln|x|

Let us start with the homogeneous equation;

x2d2ydx2+xdydx=0

Now you can realise this is the Euler Cauchy equation which shows the transformation;

x2d2ydx2+bxdydx+cy=0

Using x=etTransforms into;

d2ydt2+(b−1)dydt+cy=0

And this equation has the auxiliary equation as;

r2+(b−1)r+c=0

And after solving this equation, and substituting back the three possible types of solutions are;

y=⎧⎩⎨⎪⎪c1|x|R1+c2|x|r2,|x|r(c1+c2ln|x|),|x|a(c1cos(bln|x|)+c2sin(bln|x|)),if the auxiliary equation as TWO distinct real roots r1,r2if the auxiliary equation has two equal roots r,if the auxiliary equation has two complex roots a±bi⎫⎭⎬⎪⎪

So our equation;

x2d2ydx2+xdydx=0 becomes;

d2ydt2+(1−1)dydt=0

d2ydt2=0

The auxiliary equation is;

r2=0→r=0

So the SOLUTION to the ORIGINAL homogeneous equation is;

y=|x|0(c1+c2ln|x|)

y=c1+c2ln|x|, this is your complementary function.

Now let's look for the particular integral for;

x2d2ydx2+xdydx=12ln|x|

Let y=a(ln|x|)b

dydx=ab(ln|x|)b−1x

d2ydx2=ab(x(b−1)(ln|x|)b−2x−(ln|x|)b−1x2)

Replacing this in our differential equation;

ab(b−1)(ln|x|)b−2−ab(ln|x|)b−1+ab(ln|x|)b−1=12ln|x|

ab(b−1)(ln|x|)b−2=12ln|x|

b−2=1→b=3

ab(b−1)=12→a=2

So particular integral is;

2(ln|x|)3

And so the solution to the second order differential equation d2ydx2+1xdydx=12ln|x| is ;

y=c1+c2ln|x|+2(ln|x|)3

Explanation: HERE U GO HAVE A GOOD DAY MATE!!