match solutions and differentials equations. Note: each equation may have more than one solution. Select all that apply. I have three of them. The first:
(a)$$3y"-3y=0$$
$$y=e^{x}$$,$$y=x^{3}$$,$$y=e^{-x}$$, or $$y=x^{-2}$$
I think its easier for me to write a our like this: $$3\frac{d^{2}y}{dx^{2}}-3y=0$$

Question

match solutions and differentials equations. Note: each equation may have more than one solution. Select all that apply. I have three of them. The first:
(a)$$3y"-3y=0$$
$$y=e^{x}$$,$$y=x^{3}$$,$$y=e^{-x}$$, or $$y=x^{-2}$$
I think its easier for me to write a our like this: $$3\frac{d^{2}y}{dx^{2}}-3y=0$$

kkutie7 · Answer

Also I want help not answers =)

lochana · Answer

I was reading about this a moment ago:). 
In homogeneous differential equations, we have this way of thinking as a quadratic equation. so what we need here is to make the differential equation into quadratic equation

lochana · Answer

assume $ \frac{d^2y}{dx^2} = m^2 \ \frac{dy}{dx} = m^1 \ y = m^0 = 1 $

kkutie7 · Answer

so I have to do integrals then to get it roughly to $$y=ax^{2}+bx+c$$

lochana · Answer

mm.. not really. we need to take roots from converted equation. And take solutions. Solutions are going to be different with the nature of roots (complex, non-complex etc)

lochana · Answer

http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

lochana · Answer

I can solve this with steps. Do you want to see it?

kkutie7 · Answer

yes I think I need to see it because nothing is clicking

superdavesuper · Answer

dont mean to butt in....but since the Q has provided choices for ans, wouldnt it be easier to simply differentiate twice each choice n substitute back into the DE to see if it works?

lochana · Answer

yes. that's right. I was trying to solve it in general way.:). you can do that too

superdavesuper · Answer

@lochana u can go ahead n show the general way :)

im sure @Kkutie7 can do the differentiation on her own if she wanna try it!

lochana · Answer

so. do substitution,  you get $3m^2 - 3 = 0 \ m^2 -1 \ (m-1)(m+1)\ = 0 \ m1= 1\  and\ m2 = -1 $, as you  can see here, roots are real numbers. that means the solution should be in the form of $y = Ae^{m1x} + Be^{m2x} \ y = Ae^{-x}+Be^{x} $

lochana · Answer

in you question, I see 2 solutions. $y = e^{x} \ and \ y =e^{-x}$

lochana · Answer

and check back substitution like @superdavesuper said

lochana · Answer

@Kkutie7 did you get it?

lochana · Answer

you can change "A" and "B" into any value. in your case, it is A = 1 ; B =1

lochana · Answer

try this http://www.intmath.com/differential-equations/7-2nd-order-de-homogeneous.php

kkutie7 · Answer

yes i got it thank you

lochana · Answer

you are welcome:)