2. Originally Posted by dapore
The reason you have not received any help is that the wording of your problem is near incomprehensible. Please consider rewording and if the original was in English repost the exact wording.

CB

3. 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.

4. Prove that the equation
$\left ( D^{2}+RD+S \right )Y=0$
has a special integration If $R+XS=0$ where $Y=X$
Then Solve the equation $\left ( D^{2}-\frac{3}{X}D+\frac{3}{X^{2}} \right )Y=2X-1$

5. Originally Posted by dapore
Prove that Of the equation

$\left ( D^{2}+RD+S \right )Y=0$
Special solution $R+XS=0$ where $Y=X$

Then Solve the equation $\left ( D^{2}-\frac{3}{X}D+\frac{3}{X^{2}} \right )Y=2X-1$
Writing your post bigger will not aid in getting you help. If anything it will probably offend most.

6. Question has been amended

7. 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.

8. Thank you my friends