Confused on simple differential equation

[BillhB]

Homework Statement

Find the values of m so that ##y = x^m## is a solution of ##x^2\frac{d^2y}{dx^2} - 3x\frac{dy}{dx} -12y = 0##

Homework Equations

##y = x^m##
##y'=mx^{m-1}##
##y''=(m^2-m)x^{m-2}##

The Attempt at a Solution

So after plugging and chugging we get

$$(m+2)(m-6)x^m = 0 $$

Henceforth m = -2 or m = 6.

All well and good but why don't we look at the case when x = 0. Then the domain of m would be ##(-\infty, 0) \cup (0, \infty)## right?

The question doesn't state we're looking for m's for any possible x, so this seems correct to me. Is it just assumed that is the case? Or am I misunderstanding something?

 

[tommyxu3]

The problem required to find the value of ##m## such that ##y=x^m## is the solution, so I think your ##m## has to make all the ##x## and ##y## fit the equation.
Besides, ##x## varies and what is given is a function. I think ##x=0## just meets a special case. (my opinion)

 

[BillhB]

The problem required to find the value of ##m## such that ##y=x^m## is the solution, so I think your ##m## has to make all the ##x## and ##y## fit the equation.
Besides, ##x## varies and what is given is a function. I think ##x=0## just meets a special case. (my opinion)

I agree but I put when X = 0 m equaled the above ... when ##X \neq 0## m = -2 or m =6 and got points deducted with a red mark through the when X = 0 part? Wondering if something about above is wrong or if I need to go to professor and seek clarification for deduction.

 

[tommyxu3]

Ummm... Then I think his viewpoints are that the value ##m\in \mathbb{R}\setminus\{0\}## cannot make all the ##x## fit the equation, in which case we cannot say ##y=x^m## is a solution to the ODE.

 

[Mark44]

Homework Statement

Find the values of m so that ##y = x^m## is a solution of ##x^2\frac{d^2y}{dx^2} - 3x\frac{dy}{dx} -12y = 0##

Homework Equations

##y = x^m##
##y'=mx^{m-1}##
##y''=(m^2-m)x^{m-2}##

The Attempt at a Solution

So after plugging and chugging we get

$$(m+2)(m-6)x^m = 0 $$

Henceforth m = -2 or m = 6.

All well and good but why don't we look at the case when x = 0. Then the domain of m would be ##(-\infty, 0) \cup (0, \infty)## right?

The goal of the exercise is to find a function y = x^m that satisfies the DE. As your work shows, two possible functions are y = x^-2 and y = x⁶.

When you substitute the equations ##y = x^m##, ##y' = mx^{m - 1}## and ##y'' = m(m - 1)x^{m - 2}## into the original diff. equation, you get ##(m + 2)(m - 6)x^m = 0##. This equation has to be true for all real x, or at least all x values in the domains of the solutions, not just for a specific x value of 0. For this reason, you don't consider the possibility that x^m = 0.

The question doesn't state we're looking for m's for any possible x, so this seems correct to me. Is it just assumed that is the case? Or am I misunderstanding something?

 

[BillhB]

The goal of the exercise is to find a function y = x^m that satisfies the DE. As your work shows, two possible functions are y = x^-2 and y = x⁶.

When you substitute the equations ##y = x^m##, ##y' = mx^{m - 1}## and ##y'' = m(m - 1)x^{m - 2}## into the original diff. equation, you get ##(m + 2)(m - 6)x^m = 0##. This equation has to be true for all real x, or at least all x values in the domains of the solutions, not just for a specific x value of 0. For this reason, you don't consider the possibility that x^m = 0.

Got you. Thanks.