What's the value of $ y^{(n)}$when $ y=\frac{x^n}{(x+1)^2(x+2)^2}$ What's the value of $\displaystyle  y^{(n)}$when $\displaystyle  y=\frac{x^n}{(x+1)^2(x+2)^2}$?
My Try:Let $\displaystyle y_n=\frac{x^n}{(x+1)^2(x+2)^2}$,so $\displaystyle y_n=xy_{n-1}$.According to Leibniz's formula,$$y_n^{(n)}=ny_{n-1}^{(n-1)}+xy_{n-1}^{(n)}$$.But I don't konw how to achieve it.

Assertion:
$$y_n^{(n)} = \frac{A_nx+B_n}{(x+1)^{n+2}}+\frac{C_nx+D_n}{(x+2)^{n+2}}$$
Now, we are going to need
$$
xy_{n}^{(n+1)}=\frac{-(n+1)A_nx^2+(A_n-B_n(n+2))x}{(x+1)^{n+3}}+\frac{-(n+1)C_nx^2+(2C_n-(n+2)D_n)x}{(x+2)^{n+3}}
$$
and
$$
(n+1)y_{n}^{(n)}=(n+1)\frac{(A_nx+B_n)(x+1)}{(x+1)^{n+3}}+(n+1)\frac{(C_nx+D_n)(x+2)}{(x+2)^{n+3}}
$$
Combining these and applying the Leibniz rule, we get
$$
y_{n+1}^{(n+1)}=\frac{((n+2)A_n-B_n)x+(n+1)B_n}{(x+1)^{n+3}}+\frac{(2(n+2)C_n-D_n)x+2(n+1)D_n}{(x+2)^{n+3}}
$$
Matching with the original assertion, we have
$$
A_{n+1}=(n+2)A_n-B_n\\
B_{n+1}=(n+1)B_n\\
C_{n+1}=2(n+2)C_n-D_n\\
D_{n+1}=2(n+1)D_n
$$
And as $y_0^{(0)} = -\frac{2x+1}{(x+1)^2}+\frac{2x+5}{(x+2)^2}$, we also have that
$$
A_0 = -2\\
B_0 = -1\\
C_0 = 2\\
D_0 = 5
$$
Now, we can solve for $B_n$ and $D_n$ relatively easily. $A_n$ and $C_n$ will require a little more work. But I'll leave the rest to you.
