Finding Values of m given a Differential equation?

Question

Okay The problem states

Find values of m so that the function y=x^m is the solution of the given differential equation.

The given differential equation is xy'' +2y'=0

Do I just find y' and y'' which would be

y'=mx^(m-1)

y''=m(m-1) * x^(m-2)

y''=(m^2 - m) * x^(m-2)

And then plug it into the equation?

Really confused about how to solve this one. Please help!

Answer 1

your methods sounds right:

You get:

x(m^2 -m)x^(m-2) + 2mx^(m-1) = 0

or (m^2 -m)x^(m-1) + 2mx^(m-1) = 0

or (m^2 + m)x^(m-1) = m(m+1)x^(m-1) = 0

Since this equation must work for every value of x (whether x=0 or not) you can only work for m = 0 or -1

Answer 2

yes.

x[m(m-1) * x^(m-2)] + 2[mx^(m-1)] = 0

x(m-1) + 2x = 0

m - 1 + 2 = 0

m = -1

y = 1/x

y ' = -1/x^2

y '' = 2/x^3

xy '' + 2y ' = x(2/x^3) + 2(-1/x^2) = 0 so the solution looks good

Answer 3

Exactly.  y' = mx^(m - 1) and y'' = (m^2 - m)x^(m-2), so plug this into the differential equation:

0 = xy'' + 2y' = x[(m^2 - m)x^(m-2)] + 2[mx^(m - 1)] = (m^2+m)[x^(m-1)]  

For the above to hold for all x, m^2+m must equal 0, which implies that m = -1 or 0.