Q 3.6-15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The solution to the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$crosses the x-axis at a point in the interval [0,14] .By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

Step-by-Step Solution

Answer

The solution on the given conditions and on the interval cross x-axis at x = 1.27.

Step 1: Find the equation of approximation value

Here $\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{x}+\mathrm{y}+2{)}^2,\mathrm{y}\left(0\right)=-2$ ,

For,  ${\mathrm{x}}_{\mathrm{o}}=0, {\mathrm{y}}_{\mathrm{o}}=-2, \text{M = 280, h = 0.005, interval = }\left[0,14\right]$

 $$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y})=(\mathrm{x}+\mathrm{y}+2{)}^2 \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =(\mathrm{x}+\mathrm{y}+2+0.005(1+(\mathrm{x}+\mathrm{y}+2{)}^2){)}^2 \end{gathered}$$

Step 2: Solve for x and y

Apply initial points $\mathrm{x}=0,\mathrm{y}=-2,\mathrm{h}=0.005$

 $$\begin{gathered} \mathrm{F}(0,-2)=0 \\ \mathrm{G}(0,-2)=0.000025 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=(\mathrm{x}+\mathrm{h}) \\ \mathrm{y}=\mathrm{x}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \\ x=0.005 \\ \mathrm{y}=-2 \end{gathered}$$

Step 3: Determine the all other values

Apply the same procedure for all other values and the values are 

 

(x = 1.270, y = -0.047)

 

(x = 1.275, y = 0.006)

 

Hence, the solution on the given conditions and on the interval cross x-axis at x=1.27.