Question 1209184
<pre>

Let y<sup>((n))</sup> indicate the nth derivative:

First we will solve the given differential equation

{{{"y'"}}}{{{""=""}}}{{{-y/x}}}

{{{dy/dx}}}{{{""=""}}}{{{-y/x}}}

We may swap means or extremes in any proportion:

{{{dy/y}}}{{{""=""}}}{{{-dx/x}}}

{{{int(dy/y)}}}{{{""=""}}}{{{-int(dx/x)}}}

{{{ln(y)}}}{{{""=""}}}{{{-ln(x)+ln(C)}}}

{{{ln(y)}}}{{{""=""}}}{{{ln(C/x)}}}

{{{y}}}{{{""=""}}}{{{C/x}}}

Using the quotient rule for derivative:

{{{"y'"}}}{{{""=""}}}{{{(x*0-C*1)/x^2}}}{{{""=""}}}{{{-C^""/x^2}}}

Using the quotient rule for derivative again:

{{{"y''"}}}{{{""=""}}}{{{-(x^2*0-C*2x)/x^4}}}{{{""=""}}}{{{2C^""/x^3}}}{{{""=""}}}{{{(2!*C^"")/x^3}}}

Using the quotient rule for derivative again:

{{{"y'''"}}}{{{""=""}}}{{{(x^3*0-2C*3x^2)/x^6}}}{{{""=""}}}{{{-3*2C^""/x^4}}}{{{""=""}}}{{{(-3!C^"")/x^4}}}

So apparently it looks as though the nth derivative is this, and since the
signs alternate, we multiply by the sign-changing factor (-1)<sup>n</sup>: 

{{{y^(((n)))}}}{{{""=""}}}{{{((-1)^n*n!*C)/x^(n+1)}}}

We can prove this inductively since we have two cases where the rule works.

Assume n=k is some case where that rule works, then

{{{y^(((k)))}}}{{{""=""}}}{{{((-1)^k*k!*C)/x^(k+1)}}}

Using the quotient rule:

{{{y^(((k+1)))}}}{{{""=""}}}{{{(x^(k+1)*0-(-1)^k*k!C*(k+1)*x^k)/x^(2(k+1))}}}{{{""=""}}}{{{((-1)^(k+1)C*k!)/x^(2k+2-k)}}}{{{""=""}}}{{{((-1)^(k+1)C*k!)/x^(k+2)}}}

which is this assumed formula for n = k+1

{{{y^(((k+1)))}}}{{{""=""}}}{{{((-1)^(k+1)*((k+1)-1)!C)/x^(k+2)}}}{{{""=""}}}{{{((-1)^(k+1)C*k!)/x^(k+2)}}}

So we have proven inductively that

{{{y^(((n)))}}}{{{""=""}}}{{{((-1)^n*n!*C)/x^(n+1)}}}

Now we show that the given formula for the nth derivative is equivalent to 
the equation we proved by substituting {{{C/x}}} for y

{{{y^(((n)))}}}{{{""=""}}}{{{((-1)^(n)*n!*y)/(x^n)}}}

{{{y^(((n)))}}}{{{""=""}}}{{{((-1)^(n)*n!(C/x))/(x^n)}}} 

Multiplying top and bottom by x

{{{y^(((n)))}}}{{{""=""}}}{{{((-1)^(n)*n!*C)/(x^(n+1))}}} 

This is the same as the equation we proved. So the claim is true.

Edwin</pre>