91Ó°ÊÓ

Use the Nonlinear Shooting Algorithm with \(h=0.5\) to approximate the solution to the boundary-value problem $$ y^{\prime \prime}=-\left(y^{\prime}\right)^{2}-y+\ln x, \quad 1 \leq x \leq 2, \quad y(1)=0, \quad y(2)=\ln 2 $$ Compare your results to the actual solution \(y=\ln x\).

Use the Linear Shooting method to approximate the solution to the following boundary-value problems. a. \(\quad y^{\prime \prime}=-3 y^{\prime}+2 y+2 x+3, \quad 0 \leq x \leq 1, y(0)=2, y(1)=1\); use \(h=0.1\). b. \(\quad y^{\prime \prime}=-4 x^{-1} y^{\prime}-2 x^{-2} y+2 x^{-2} \ln x, \quad 1 \leq x \leq 2, y(1)=-\frac{1}{2}, y(2)=\ln 2 ;\) use \(h=0.05\). c. \(\quad y^{\prime \prime}=-(x+1) y^{\prime}+2 y+\left(1-x^{2}\right) e^{-x}, \quad 0 \leq x \leq 1, y(0)=-1, y(1)=0\); use \(h=0.1\). d. \(\quad y^{\prime \prime}=x-1 y^{\prime}+3 x^{-2} y+x^{-1} \ln x-1, \quad 1 \leq x \leq 2, y(1)=y(2)=0\); use \(h=0.1\).

Use the Linear Finite-Difference Algorithm to approximate the solution to the following boundaryvalue problems. a. \(\quad y^{\prime \prime}=-3 y^{\prime}+2 y+2 x+3, \quad 0 \leq x \leq 1, y(0)=2, y(1)=1 ;\) use \(h=0.1\). b. \(\quad y^{\prime \prime}=-4 x^{-1} y^{\prime}+2 x^{-2} y-2 x^{-2} \ln x, \quad 1 \leq x \leq 2, y(1)=-\frac{1}{2}, y(2)=\ln 2 ;\) use \(h=0.05\). c. \(\quad y^{\prime \prime}=-(x+1) y^{\prime}+2 y+\left(1-x^{2}\right) e^{-x}, \quad 0 \leq x \leq 1, y(0)=-1, y(1)=0 ;\) use \(h=0.1\). d. \(\quad y^{\prime \prime}=x^{-1} y^{\prime}+3 x^{-2} y+x^{-1} \ln x-1, \quad 1 \leq x \leq 2, y(1)=y(2)=0 ;\) use \(h=0.1\).

Use the Nonlinear Finite-Difference Algorithm with \(T O L=10^{-4}\) to approximate the solution to the following boundary-value problems. The actual solution is given for comparison to your results. a. \(\quad y^{\prime \prime}=-e^{-2 y}, \quad 1 \leq x \leq 2, y(1)=0, y(2)=\ln 2 ;\) use \(N=9\); actual solution \(y(x)=\ln x\). b. \(\quad y^{\prime \prime}=y^{\prime} \cos x-y \ln y, \quad 0 \leq x \leq \frac{\pi}{2}, y(0)=1, y\left(\frac{\pi}{2}\right)=e ;\) use \(N=9 ;\) actual solution \(y(x)=e^{\sin x}\). c. \(y^{\prime \prime}=-\left(2\left(y^{\prime}\right)^{3}+y^{2} y^{\prime}\right) \sec x, \quad \frac{\pi}{4} \leq x \leq \frac{\pi}{3}, y\left(\frac{\pi}{4}\right)=2^{-1 / 4}, y\left(\frac{\pi}{3}\right)=\frac{1}{2} \sqrt[4]{12} ;\) use \(N=4\); actual solution \(y(x)=\sqrt{\sin x}\) d. \(\quad y^{\prime \prime}=\frac{1}{2}\left(1-\left(y^{\prime}\right)^{2}-y \sin x\right), \quad 0 \leq x \leq \pi, y(0)=2, y(\pi)=2 ;\) use \(N=19\); actual solution \(y(x)=2+\sin x\)

Use the Linear Finite-Difference Algorithm to approximate the solution \(y=e^{-10 x}\) to the boundaryvalue problem $$ y^{\prime \prime}=100 y, \quad 0 \leq x \leq 1, \quad y(0)=1, \quad y(1)=e^{-10} $$ Use \(h=0.1\) and \(0.05\). Can you explain the consequences?