Find all the values of r such that y=Ax^r satisfies the equation:
y''+(x^-1)y'-y^3=0

Give a reason why other solutions must exist for this equation.

Question

Find all the values of r such that y=Ax^r satisfies the equation:
y''+(x^-1)y'-y^3=0

Give a reason why other solutions must exist for this equation.

myininaya · Answer

$$y=Arx^{r-1}. y'=Ar(r-1)x^{r-2}, y''=Ar(r-1)(r-2)x^{r-3}$$
plug in

anonymous · Answer

First take the first and second derivatives of your function y with respect to x and plug them into the second order differential equation

anonymous · Answer

$$r(r-1)Ax^{r-2} + rAx^{r-2}-A^{3} x^{3r} =0$$

anonymous · Answer

Myininaya the function should be $$y=Ax^r$$

anonymous · Answer

$$y \prime= rAx ^{r-1}$$

anonymous · Answer

$$y \prime \prime = r(r-1)A x^{r-2}$$

myininaya · Answer

lol

myininaya · Answer

thats what i get for going over the speed limit

anonymous · Answer

hahaha your response was quick

myininaya · Answer

my y'' looked to nasty to plug in
i felt sorry for him
but now i don't

anonymous · Answer

Yeah

anonymous · Answer

Here it is with the values plugged in, where do I go from here?
$$r(r-1)Ax^{r-2} + rAx^{r-2}-A^{3} x^{3r} =0$$

anonymous · Answer

Well, A= 0 clearly works...:-)

anonymous · Answer

We could try adding the coefficient of $$A x^{r-2}$$

anonymous · Answer

To simplify the first two terms:

anonymous · Answer

$$A x^{r-2}[r^{2} - r + r]$$

anonymous · Answer

$$Ar ^{2} x^{r-2}$$ - $$A^{3} x^{3r} = 0$$

anonymous · Answer

Now we have to think lol... Although one of the obvious answers f

anonymous · Answer

for the other solutions will be that y = 0  is a solution