SOLUTION: if y ' = - (y)/(x) then y ^((n)) = (((- 1))^(n) n ! y)/(x ^(n)) , n \[Element] N (true or fals)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: if y ' = - (y)/(x) then y ^((n)) = (((- 1))^(n) n ! y)/(x ^(n)) , n \[Element] N (true or fals)      Log On


   



Question 1209184: if y ' = - (y)/(x) then y ^((n)) = (((- 1))^(n) n ! y)/(x ^(n)) , n \[Element] N (true or fals)
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

Let y((n)) indicate the nth derivative:

First we will solve the given differential equation

%22y%27%22%22%22=%22%22-y%2Fx

dy%2Fdx%22%22=%22%22-y%2Fx

We may swap means or extremes in any proportion:

dy%2Fy%22%22=%22%22-dx%2Fx

int%28dy%2Fy%29%22%22=%22%22-int%28dx%2Fx%29

ln%28y%29%22%22=%22%22-ln%28x%29%2Bln%28C%29

ln%28y%29%22%22=%22%22ln%28C%2Fx%29

y%22%22=%22%22C%2Fx

Using the quotient rule for derivative:

%22y%27%22%22%22=%22%22%28x%2A0-C%2A1%29%2Fx%5E2%22%22=%22%22-C%5E%22%22%2Fx%5E2

Using the quotient rule for derivative again:

%22y%27%27%22%22%22=%22%22-%28x%5E2%2A0-C%2A2x%29%2Fx%5E4%22%22=%22%222C%5E%22%22%2Fx%5E3%22%22=%22%22%282%21%2AC%5E%22%22%29%2Fx%5E3

Using the quotient rule for derivative again:

%22y%27%27%27%22%22%22=%22%22%28x%5E3%2A0-2C%2A3x%5E2%29%2Fx%5E6%22%22=%22%22-3%2A2C%5E%22%22%2Fx%5E4%22%22=%22%22%28-3%21C%5E%22%22%29%2Fx%5E4

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

y%5E%28%28%28n%29%29%29%22%22=%22%22%28%28-1%29%5En%2An%21%2AC%29%2Fx%5E%28n%2B1%29

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%5E%28%28%28k%29%29%29%22%22=%22%22%28%28-1%29%5Ek%2Ak%21%2AC%29%2Fx%5E%28k%2B1%29

Using the quotient rule:

y%5E%28%28%28k%2B1%29%29%29%22%22=%22%22%22%22=%22%22%28%28-1%29%5E%28k%2B1%29C%2Ak%21%29%2Fx%5E%282k%2B2-k%29%22%22=%22%22%28%28-1%29%5E%28k%2B1%29C%2Ak%21%29%2Fx%5E%28k%2B2%29

which is this assumed formula for n = k+1

y%5E%28%28%28k%2B1%29%29%29%22%22=%22%22%28%28-1%29%5E%28k%2B1%29%2A%28%28k%2B1%29-1%29%21C%29%2Fx%5E%28k%2B2%29%22%22=%22%22%28%28-1%29%5E%28k%2B1%29C%2Ak%21%29%2Fx%5E%28k%2B2%29

So we have proven inductively that

y%5E%28%28%28n%29%29%29%22%22=%22%22%28%28-1%29%5En%2An%21%2AC%29%2Fx%5E%28n%2B1%29

Now we show that the given formula for the nth derivative is equivalent to 
the equation we proved by substituting C%2Fx for y

y%5E%28%28%28n%29%29%29%22%22=%22%22%28%28-1%29%5E%28n%29%2An%21%2Ay%29%2F%28x%5En%29

y%5E%28%28%28n%29%29%29%22%22=%22%22%28%28-1%29%5E%28n%29%2An%21%28C%2Fx%29%29%2F%28x%5En%29 

Multiplying top and bottom by x

y%5E%28%28%28n%29%29%29%22%22=%22%22%28%28-1%29%5E%28n%29%2An%21%2AC%29%2F%28x%5E%28n%2B1%29%29 

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

Edwin