Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$ 
Question: Given that $X$ and $Y$ are two random variables  satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$ 
  Calculate 
  $$\mathbb{E}(X-Y \mid 2X+Y).$$

I tried to use the fact that if $A$ and $B$ are independent, then $\mathbb{E}(A\mid B) = \mathbb{E}(A)$ and uncorrelated implies independence in jointly normal distribution. 
So, I attempted to express $X-Y$ as a linear combination of $2X+Y$ and $Z$ where $\operatorname{Cov}(2X+Y,Z) = 0.$
But I am not able to do so.
Any hint is appreciated.

We use two property:
First: $E(2X+Y|2X+Y)=2X+Y$ 
Second:  $(X-dY,2X+Y)$ is bi-variate normal(for $d\neq - \frac{1}{2}$), if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent(by set $\rho=0$ in bivarite distribution of joint $(X-dY,2X+Y)$
Correlations_and_independence). so $E(X-dY|2X+Y)=E(X-dY)=0$.
$$E(2X+Y|2X+Y)=2X+Y$$
so 
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y)  \hspace{1cm} (1)$$
For first step let $\rho=0$
$$cou(X-2\frac{a}{b} Y,2X+Y)=2Var(X)-2\frac{a}{b} Var(Y)=2a-2\frac{a}{b}b=0$$
so since $X-2\frac{a}{b} Y$ and $2X+Y$$ are normal so they are independent.
in hence 
$$E(X-2\frac{a}{b} Y|2X+Y)=E(X-2\frac{a}{b} Y)=0$$
$$E(X|2X+Y)=2\frac{a}{b} E(Y|2X+Y)\hspace{1cm} (2)$$
combine (1) and (2)
$$E(X|2X+Y)=\frac{2\frac{a}{b}}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$ 
$$E(Y|2X+Y)=\frac{1}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$ 
so 
$$E(X -Y|2X+Y)=(\frac{2\frac{a}{b}}{1+4\frac{a}{b}}-\frac{1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)=(\frac{2\frac{a}{b}-1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)$$
**Now for general case ** $\rho \in[-1,1]$
if $$d=\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}} \hspace{1cm} (3)$$
$$cou(X-dY,2X+Y)=2a-db+(1-2d)\rho \sqrt{a} \sqrt{b}$$
$$=2a+\rho \sqrt{a} \sqrt{b}-d(b+2\rho \sqrt{a} \sqrt{b})=0$$
so $$E(X-dY|2X+Y)=E(X-dY)=0$$ and in hence 
$$E(X|2X+Y)=dE(Y|2X+Y) \hspace{1cm} (4)$$
Combine (4) and (1) 
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y)=2X+Y-2dE(Y|2X+Y)$$ so
$$E(Y|2X+Y)=\frac{1}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (5)$$ and 
$$E(X|2X+Y)=dE(Y|2X+Y)=\frac{d}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (6)$$
(5) and (6)
$$E(X-Y|2X+Y)=\frac{d-1}{1+2d}\bigg(2X+Y\bigg)$$
$$=\frac{\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}-1}{1+2\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}}\bigg(2X+Y\bigg)$$
$$=\frac{2a-b-\rho \sqrt{a} \sqrt{b}}{b+4a+4\rho \sqrt{a} \sqrt{b}}\bigg(2X+Y\bigg)$$
detail for  "@Student" 
I explain now  why I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent. 
1)$(X-dY,2X+Y)$ is bi-variate normal for $d\neq \frac{-1}{2}$
I can write
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y 
\end{bmatrix}
=\begin{bmatrix}
1 & -d \\
2 & 1 
\end{bmatrix}
\begin{bmatrix}
X \\
Y 
\end{bmatrix}
\end{eqnarray}
By linear-transformation-of-gaussian-random-variable I think 
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y 
\end{bmatrix}
\end{eqnarray}
is bi-variate normal.
2) Now by Correlations_and_independence I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent. wikipedia: "In general, random variables may be uncorrelated but statistically dependent. But if a random vector has a multivariate normal distribution then any two or more of its components that are uncorrelated are independent".
