Question

The solution to the initial value problem$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$has a vertical asymptote (“blows up”) at some point in the interval [1,2]By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

Short answer

The solution on the given conditions and on the interval (“blows up”) at x = 1.26.

Step-by-step solution

Find the equation of approximation value

Here$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^3{\mathrm{y}}^2,\mathrm{y}\left(1\right)=3$,

For, ${\mathrm{x}}_{\mathrm{o}}=1,{\mathrm{y}}_{\mathrm{o}}=3,\text{M = 280, h = 0.005, interval =}\left[1,2\right]$

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y})=\left({\mathrm{x}}^3{\mathrm{y}}^2-\frac{\mathrm{y}}{\mathrm{x}}\right) \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =(\mathrm{x}+0.005{)}^3{\left(\mathrm{y}+0.005\left({\mathrm{x}}^3{\mathrm{y}}^2-\frac{\mathrm{y}}{\mathrm{x}}\right)\right)}^2-\frac{\mathrm{y}+0.005\left({\mathrm{x}}^3{\mathrm{y}}^2-\frac{\mathrm{y}}{\mathrm{x}}\right)}{\mathrm{x}+0.005} \end{gathered}$$

Solve for x and y

Apply initial points $\mathrm{x}=0,\mathrm{y}=3,\mathrm{h}=0.005$

$$\begin{gathered} \mathrm{F}(1,3)=6 \\ \mathrm{G}(1,3)=6.030 \\ \mathrm{x}=({\mathrm{x}}_{\mathrm{o}}+\mathrm{h}) \\ \mathrm{y}={\mathrm{y}}_{\mathrm{o}}+\frac{\mathrm{h}}{2}(\mathrm{F}+\mathrm{G}) \\ \mathrm{x}=0.005 \\ \mathrm{y}=3 \end{gathered}$$

Determine the all-other values

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

(x = 1.261, y = 2197….)

(x = 1.262, y = 11800…)

Hence, the solution on the given conditions and on the interval cross x-axis at x = 1.26.