(a) Show that ${\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0}$ is an implicit solution to$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}\mathbf{y}}$ on the interval $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{3}\mathbf{)}$.

(b) Show that${\mathbf{xy}}^{\mathbf{3}}\mathbf{-}{\mathbf{xy}}^{\mathbf{3}}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{1}$ is an implicit solution to$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\left(\mathbf{x}\mathbf{cos}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{x}\mathbf{-}\mathbf{1}\right)\mathbf{y}}{\mathbf{3}\left(\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{sin}\mathbf{x}\right)}$ on the interval $\left(0,\frac{\pi }{2}\right)$.

Firstly, we will differentiate${\mathrm{y}}^2+\mathrm{x}-3=0$,with respect tox,

$2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}+1=0$

After differentiating, we will simplify the differential equation we got, to get the given differential equation.

$$\begin{gathered} 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=-1 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2\mathrm{y}} \end{gathered}$$

Which is identical to the given differential equation.

Hence, ${\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0}$ is an implicit solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}\mathbf{y}}$ on the interval $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{3}\mathbf{)}$.

Firstly, we will differentiate ${\mathrm{xy}}^3-{\mathrm{xy}}^3\sin \mathrm{x}=1$, with respect tox,

$$\begin{gathered} {\mathrm{xy}}^3\times \left(-\cos \mathrm{x}\right)+\left(1-\sin \mathrm{x}\right)\frac{\mathrm{d}\left({\mathrm{xy}}^3\right)}{\mathrm{dx}}=0 \\ -{\mathrm{xy}}^3\times \cos \mathrm{x}+\left(1-\sin \mathrm{x}\right)\left({\mathrm{y}}^3+3{\mathrm{xy}}^2\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \end{gathered}$$

After differentiating, we will simplify the differential equation we got, to get the given differential equation.

$$\begin{gathered} {\mathrm{y}}^3+3{\mathrm{xy}}^2\frac{\mathrm{dy}}{\mathrm{dx}}-3{\mathrm{xy}}^2\sin \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-{\mathrm{y}}^3\sin \mathrm{x}={\mathrm{xy}}^3\cos \mathrm{x} \\ 3{\mathrm{y}}^2\left(\mathrm{x}-\mathrm{x}\sin \mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{xy}}^3\cos \mathrm{x}-{\mathrm{y}}^3+{\mathrm{y}}^3\sin \mathrm{x} \\ 3{\mathrm{y}}^2\left(\mathrm{x}-\mathrm{x}\sin \mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{x}\cos \mathrm{x}+\sin \mathrm{x}-1\right){\mathrm{y}}^3 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\mathrm{x}\cos \mathrm{x}+\sin \mathrm{x}-1\right)\mathrm{y}}{3\left(\mathrm{x}-\mathrm{x}\sin \mathrm{x}\right)} \end{gathered}$$

Which is identical to the given differential equation.

Therefore,${\mathbf{xy}}^{\mathbf{3}}\mathbf{-}{\mathbf{xy}}^{\mathbf{3}}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{1}$ is an implicit solution to$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\left(\mathbf{x}\mathbf{cos}\mathbf{x}\mathbf{-}\mathbf{sin}\mathbf{x}\mathbf{-}\mathbf{1}\right)\mathbf{y}}{\mathbf{3}\left(\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{sin}\mathbf{x}\right)}$ on the interval$\left(0,\frac{\pi }{2}\right)$