Please I need a solution

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- Dec 21st 2009, 07:30 AM #1

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- Dec 21st 2009, 07:39 PM #2

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- Dec 23rd 2009, 04:37 AM #3

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There is no

**predicate**in

"Prove that the integration $\displaystyle (D^2+ RD+ S)Y= 0$

especially where y= x if R+ SX= 0"

Prove the the integeration is**what**? And it's not clear what "R+ SX= 0" means. Is "X" the same as "x"? If not, what is it? If so, R+ Sx= 0 is true for just one value of x, specifically x= -R/S.

The second equation can be written as $\displaystyle X^2D^2- 3XD+ 3)Y= 0$ by multiplying both sides by $\displaystyle X^2$. And then the change of variable x= ln(t) gives an equation "with constant coefficients" like the first equation.

- Dec 24th 2009, 02:25 PM #4

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- Dec 24th 2009, 03:02 PM #5

- Dec 24th 2009, 03:08 PM #6

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- Dec 26th 2009, 03:11 AM #7

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You are now saying, since "X= Y", that you must prove that R+ YS= 0, which is the same as Y= -R/S, satisfies the equation $\displaystyle (D^2+ RD+ S)Y= 0$.

That makes sense now, but is wrong! If you replace Y by the**constant**function -R/S in that equation, $\displaystyle D^2Y$ and DY are both 0 since Y is a constant and the equation reduces to SY= -R= 0 which is not generally true.

Perhaps you meant "Show that Y= R+ XS, where X is the independent variable, satisfies $\displaystyle (D^2+ RD+ S)Y= 0$."

Now, Dy= D(R+ XS)= S and $\displaystyle D^2Y= D^2(R+ XS)= D(S)= 0$ so the equation becomes 0+ R(S)+ (R+XS)(S) which is still not 0.

I still have no idea what you are really asking.

- Dec 30th 2009, 01:22 AM #8