Question

Transmission Lines.In the study of the electric field that is induced by two nearby transmission lines, an equation of the form$\frac{\mathit{d}\mathit{z}}{\mathit{d}\mathit{x}}\mathbf{+}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$arises. Let$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{2}$and$\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$. If z(0)=1, use the fourth-order Runge–Kutta algorithm to approximate z(1). For a tolerance of$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0001}$, use a stopping procedure based on the absolute error.

Short answer

$z\left(1\right)=2.870833$with $h=0.03125$

Step-by-step solution

Find the values of ki.i=1,2,3,4

Using the improved 4^th order Runge-Kutta subroutine with tolerance $\xi =0.0001$.

Since $f(x,z)=5x+2-x^2{z}^2$and $x=x_0=0,z=z_o=1$and $h=1$

$$\begin{gathered} k_1=hf(x,z)=h(5x+2-x^2{z}^2) \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=h\left[5\left(x+\frac{h}{2}\right)+2-{\left(x+\frac{h}{2}\right)}^2{\left(z+\frac{k_1}{2}\right)}^2\right] \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=h\left[5\left(x+\frac{h}{2}\right)+2-{\left(x+\frac{h}{2}\right)}^2{\left(z+\frac{k_2}{2}\right)}^2\right] \\ {k}_4=hf\left(x+h,z+k_3\right)=h\left[5(x+h)+2(x+h{)}^2{(z+k_3)}^2\right] \\ {k}_1=h(x,z)=2 \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=3.5 \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=2.60938 \\ {k}_4=hf\left(x+h,z+k_3\right)=0.-6.02759 \end{gathered}$$

Find the values of x and z

$$\begin{gathered} x=x_o+h=1 \\ z=z_o+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =2.365194 \end{gathered}$$

Therefore $z\left(1\right)=2.465194$.

This is the solution of IVP.

Now,

role="math" localid="1664088788213" $$\begin{gathered} \xi =\left|\frac{1-2.365194}{1}\right| \\ \xi =1.365194>0.0001 \end{gathered}$$

Determine the other values

Apply the same procedure for $h=0.5$, $x=0,z=1$.

$$\begin{gathered} k_1=hf(x,z)=1 \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=1.55469 \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=1.52628 \\ {k}_4=hf\left(x+h,z+k_3\right)=1.45224 \\ x=0+0.5=0.5 \\ z=1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =2.4357 \end{gathered}$$

Thus,$z\left(1\right)=2.573986$

Now

$\xi =\left|2.365194-2.573986\right|=0.208792>0.0001$

Evaluate the other values

Apply the same procedure for other values. The values are

$$\begin{gathered} f\left(1\right)=v(1;0.25)=2.854533 \\ \xi =\left|2.573986-2.854553\right|=0.280574>0.0001 \\ f\left(1\right)=v(1;0.125)=2.870202 \\ \xi =\left|2.870202-2.854553\right|=0.015649>0.0001 \\ f\left(1\right)=v(1;0.0625)=2.870805 \\ \xi =\left|2.870805-2.870202\right|=0.000603>0.0001 \\ f\left(1\right)=v(1;0.03125)=2.870833 \\ \xi =\left|2.870833-2.870805\right|=0.000028<0.0001 \end{gathered}$$

Hence, $z\left(1\right)=2.870833$.

Hence the solution is $z\left(1\right)=2.870833$with $h=0.03125$