<s>[INST] Use the method of "changing variables" to solve the following recurrence: $T(n) = 2\cdot T(n-2) + n$
I tried doing this, but I don't know how to continue from here. I think it doesn't work and is not correct.

Please help, thanks in advance!
 [/INST] Let $f(x)=T_0+T_1x+T_2x^2+....+T_nx^n+...$. Thus, $x^2f=T_0x^2+T_1x^3+T_2x^4+...+T_{n-2}x^n+....$. Also, note that $\frac{x}{(1-x)^2}=x+2x^2+3x^3+...nx^n+...$
Thus,
$$f-2x^2f-\frac{x}{(1-x)^2}=T_0+(T_1-1)x+(T_2-2T_0-2)x^2+...+(T_n-2T_{n-2}-n)x^n+...$$
$$=T_0+(T_1-1)x$$
Thus,
$$f=\frac{T_0}{1-2x^2}+\frac{x}{(1-2x^2)(1-x)^2}+\frac{(T_1-1)x}{1-2x^2}$$
Now, $\frac{x}{(1-2x^2)(1-x)^2}= \frac{3}{\left(x-1 \right)}+{\frac {-6\,x-4}{2\,{x}^{2}-1}}- \frac{1}{\left( 
x-1 \right) ^{2}}$
Thus atlast we have:
$$f=\frac{T_0+4}{1-2x^2}+\frac{(T_1+5)x}{1-2x^2}+\frac{3}{x-1}-\frac{1}{(x-1)^2}$$
All the above functions are closed forms of well knows geometric series, from which we can find the coefficient of $x^n$ which gives us $T_n$(From the definition of $f$ in the beginning of the answer)
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