The first step is to obtain a self-contained recursion. By the $1^{st}$ recursion $x_{2k+1}=x_{2k}+x_{2k-1}$, we have $$x_{2k}=x_{2k+1}-x_{2k-1}\implies x_{2k-2}=x_{2k-1}-x_{2k-3}$$ Substituting $x_{2k}$ and $x_{2k-2}$ in the $2^{nd}$ recursion using these two relations leads to $$x_{2k+1}-x_{2k-1}=x_{2k-1}+2(x_{2k-1}-x_{2k-3})\implies x_{2k+1}=4x_{2k-1}-2x_{2k-3}$$ Letting $y_k = x_{2k-1}$ implies $$y_{k+1}=4y_k-2y_{k-1}$$ whose characteristic equation and its roots are $$t^2 = 4t-2 \implies t_{1,2}=2\pm\sqrt{2}$$ Hence, the solution for $y_k$ is $$y_k=C_1(2+\sqrt{2})^{k-1} +C_2(2-\sqrt{2})^{k-1} $$ Setting $y_1=x_1 = 2$ and $y_2=x_3=x_2+x_1=5$ to the above solution gives: $$C_{1,2}=\frac{4\pm\sqrt{2}}{4}$$ or $$x_{2k-1}=y_k=\frac{4+\sqrt{2}}{4}\times(2+\sqrt{2})^{k-1}+\frac{4-\sqrt{2}}{4}\times(2-\sqrt{2})^{k-1}$$ \indent It follows that \begin{align*} x_{2k} &= x_{2k+1}-x_{2k-1}\\ &= \Big(\frac{4+\sqrt{2}}{4}\times(2+\sqrt{2})^{k}+\frac{4-\sqrt{2}}{4}\times(2-\sqrt{2})^{k}\Big)\\ &- \Big(\frac{4+\sqrt{2}}{4}\times(2+\sqrt{2})^{k-1}+\frac{4-\sqrt{2}}{4}\times(2-\sqrt{2})^{k-1}\Big)\\ &=\frac{2\sqrt{2}+1}{4}\times(2+\sqrt{2})^k - \frac{2\sqrt{2}-1}{4}\times(2-\sqrt{2})^k \end{align*} Therefore the desired answer is $$ x_n= \left\{ \begin{array}{ll} \frac{2\sqrt{2}+1}{4}\times(2+\sqrt{2})^k - \frac{2\sqrt{2}-1}{4}\times(2-\sqrt{2})^k&\quad (n=2k)\\ \\ \frac{4+\sqrt{2}}{4}\times(2+\sqrt{2})^{k-1}+\frac{4-\sqrt{2}}{4}\times(2-\sqrt{2})^{k-1}&\quad (n=2k-1) \end{array} \right. $$ where $k = 1, 2, 3, \cdots$.