I have a problem with this exercise.
Let $f : R^{1+2} \to R^2$ be given as $f(t,(x_1,x_2)) = (x_2,x_1)$.
(a) Show through abstract arguments that there exists a unique, global solution $h: R \to R^2$ to the initial value problem $y'(t) = f(t,y(t)),y(0)=(1,0)$.
(b) Let $h = (h_1,h_2)$ be the solution from (a). Does there exist $a,b \in R$ such that $ah_1(t) + bh_2(t) = e^t$ or/and $ah_1(t) + bh_2(t) = \cos(t)$?
It is mainly part (b) which is troubling me. My immediate thought about the solution to the question, is that $h$ could maybe be defined as $h(t) = (cos(t),sin(t))$ or $h(t) = (e^t, te^t)$, but then a don't know what the constants would do, since $h$ would just solve the IVP with constants $a=b=1$ - Do you think this could be the case?
As for part (a) - it's enough to check that $f$ is Lipschitz on $\mathbb{R}\times\mathbb{R}^2$, which is easy: $$|f(t,x)-f(t,y)|=\left|\binom{x_2}{x_1}- \binom{y_2}{y_1}\right| =\left|\binom{x_1}{x_2}- \binom{y_1}{y_2}\right| =|x-y|.$$
Now for part (b). You can solve the IVP by noting that $y_1''(t)=y_1$ and $y_2''(t)=y_2$. Inserting the general solutions of those ODE's we can restrict the constants for our system. In this manner we arrive at that conclusion that: $$h_1(t)=\frac12 e^t+\frac12e^{-t}$$ $$h_1(t)=\frac12 e^t-\frac12e^{-t}$$
Now choosing $a=b=1$ we can ensure that the first condition is true. But no linear combination of $h_1$ and $h_2$ gives $\cos(t)$ - thus the second condition can never be attained.