Consider the space $X=l^2$ of the square summable sequences $u=(u_1,u_2,...)$ where $$\vert\vert u \vert\vert _{l^2}:=(\sum_{k=1}^\infty \vert u_k \vert^2)^\frac{1}{2} < \infty$$ and the function $f$ on the unit ball $\overline B(0,1) \subset l^2$ defined by $$f(u):=(\sum_{ k=1}^\infty \frac{2^{-k}}{1+3^{-k}-u_k},0,0,...). $$
How can I solve the initial value problem $$u'(t)=f(u(t)), u_1(0)=2, u_2(0)=\frac{19}{9}, u_k(0)=3^{-k}$$ for $k \geq 3$ on a suitable time interval $I$? I guess the approach is to separate the variables but I don"t know how to start with the separation.. Also, is $u \in C^1(I,l^2)$?
From $$u'(t)=f(u(t)), u_1(0)=2, u_2(0)=\frac{19}{9}, u_k(0)=3^{-k}$$ one has the following system of differential equations \begin{eqnarray} &&u_1'(t)=\sum_{ k=1}^\infty \frac{2^{-k}}{1+3^{-k}-u_k}, u_1(0)=2,\tag{1}\\ &&u_2'(t)=0,u_2(0)=\frac{19}{9},\tag{2}\\ &&u_k'(t)=0,u_k(0)=3^{-k},k=3,4,5,\cdots.\tag{3} \end{eqnarray} From (2), $u_2(t)=\frac{19}{9}$ and from (3), $u_k(t)=3^{-k}$ for $k=3,4,5,\cdots$. So (1) becomes $$ u_1'(t)=\frac{2^{-1}}{1+3^{-1}-u_1}+\frac{2^{-2}}{1+3^{-2}-\frac{19}{9}}+\sum_{k=3}^\infty 2^{-k}, u_1(0)=2$$ or $$ u_1'(t)=\frac{3}{8-6u_1(t)},u_1(0)=2$$ whose solution is $$ u_1(t)=\frac{1}{3}(4+\sqrt{4-9t}).$$ Thus $$ u(t)=\{\frac{1}{3}(4+\sqrt{4-9t}), \frac{19}{9},3^{-3},3^{-4},\cdots\}.$$