The directional field for $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{4}\mathbf{x}}{\mathbf{y}}$in shown in figure 1.12.

(a) Verify that the straight lines $\mathbf{y}\mathbf{=}\mathbf{\pm }\mathbf{2}\mathbf{x}$are solution curves, provided $\mathbf{x}\mathbf{\ne }\mathbf{0}$.

(b) Sketch the solution curve with initial condition y (0) = 2.

(c) Sketch the solution curve with initial condition y(2) = 1.

(d) What can you say about the behaviour of the above solution as $\mathbf{x}\mathbf{\to }\mathbf{+}\mathbf{\infty }$? How about $\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{\infty }$?

As $\mathrm{y}=\pm 2\mathrm{x}$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\pm 2\\ \frac{4\mathrm{x}}{\mathrm{y}}=\pm 2 (\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{x}}{\mathrm{y}})\\ \frac{4\mathrm{x}}{\pm 2x}=\pm 2 (\mathrm{x}\ne 0)\end{array}$

This $\mathbf{y}\mathbf{=}\mathbf{\pm }\mathbf{2}\mathbf{x}$is the solution curve of $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{4}\mathbf{x}}{\mathbf{y}}$for any interval except x ≠ 0.

The curve is $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{x}}{\mathrm{y}}$

$\mathrm{ydy}=4\mathrm{xdx}$

Integrating on both sides

$\begin{array}{l}\frac{{\mathrm{y}}^2}{2}=4(\frac{{\mathrm{x}}^2}{2})+\mathrm{c}\\ \frac{{\mathrm{y}}^2}{2}=2{\mathrm{x}}^2+\mathrm{c} .....\left(1\right)\end{array}$

Put the point (0, 2) in equation (1)

$\begin{array}{l}\frac{{\left(2\right)}^2}{2}=2(0{)}^2+\mathrm{c}\\ \mathrm{c}=2\\ \frac{{\mathrm{y}}^2}{2}=2{\mathrm{x}}^2+2\end{array}$

Hence this is the solution curve with initial condition y (0) = 2.

The curve is$\frac{{\mathrm{y}}^2}{2}=2{\mathrm{x}}^2+\mathrm{c}.....\left(2\right)$

Put the value of (2, 1) in equation (2)

$\begin{array}{l}\frac{{\left(1\right)}^2}{2}=2(2{)}^2+\mathrm{c}\\ 8+\mathrm{c}=\frac{1}{2}\\ c=\frac{15}{2}\\ \frac{{\mathrm{y}}^2}{2}=2{\mathrm{x}}^2+\frac{15}{2}\end{array}$

Hence this is the solution curve with initial condition y(2) = 1.

On solving parts (b) and (c), we get an increased unboundedly curve and have the lines

y = 2x and$\mathrm{y}=\pm 2\mathrm{x}$ as slant asymptotes as$\mathrm{x}\to \infty \mathrm{or}\mathrm{x}\to -\infty$. And in part (c) also increases without bounds as$\mathrm{x}\to \infty$and approaches to line y = 2x and not for $\mathrm{x}<0$.

Therefore the curves are increasing and unbounded when$\mathrm{x}\to \infty \mathrm{or}\mathrm{x}\to -\infty$.

Hence, the solutions in parts (b) and (c) become infinite and have the line

y = 2x as an asymptote $\mathbf{x}\mathbf{\to }\mathbf{\infty }$. As $\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{\infty }$, the solution in part (b) becomes infinite and has y = -2x as an asymptote, the solution in part (c) does not even exist for x negative.