91Ó°ÊÓ

Prove that when the fourth-order Runge-Kutta method is applied to the problem \(x^{\prime}=\lambda x\), the formula for advancing this solution will be $$ x(t+h)=\left[1+h \lambda+\frac{1}{2} h^{2} \lambda^{2}+\frac{1}{6} h^{3} \lambda^{3}+\frac{1}{24} h^{4} \lambda^{4}\right] x(t) $$

Consider the two-point boundary-value problem $$ \left\\{\begin{array}{l} x^{\prime \prime}=f(t, x) \\ x(0)=x(1)=0 \end{array}\right. $$ For which function(s) \(f\) can we be sure that a unique solution exists? i. \(f(t, x)=t^{2}\left(1+x^{2}\right)^{-1}\) ii. \(f(t, x)=t \sin x\) iii. \(f(t, x)=(\tan t)(\tan x)\) iv. \(f(t, x)=t / x\) v. \(f(t, x)=x^{1 / 3}\)