Question: In Problems 1-30, solve the equation.

$\left[\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{cos}\left(\mathrm{xy}\right)\right]\mathbf{dx}\mathbf{+}\left[\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)\right]\mathbf{dy}\mathbf{=}\mathbf{0}$

Given that, $\left[\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{xycos}\left(\mathrm{xy}\right)\right]\mathbf{dx}\mathbf{+}\left[\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)\right]\mathbf{dy}\mathbf{=}\mathbf{0}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let us check whether the given equation is exact or not.

Then, $\mathbf{M}\mathbf{=}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\mathbf{\partial }\mathbf{M}}{\mathbf{\partial }\mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{-}{\mathbf{x}}^2\mathbf{y}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{a}\\ \frac{\mathbf{\partial }\mathbf{N}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{-}{\mathbf{x}}^2\mathbf{y}\mathbf{sin}\left(\mathrm{xy}\right)\end{array}$

So, the given equation is exact.

Now, let us assume$\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)$

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)\right)\mathbf{dx}\\ \mathbf{=}\int \mathbf{sin}\left(\mathrm{xy}\right)\mathbf{dx}\mathbf{+}\int \mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{dx}\end{array}$

Find the value of$\int \mathbf{sin}\left(\mathrm{xy}\right)\mathbf{dx}$and$\int \mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{dx}$separately.

Case (1):

$\int \mathbf{sin}\left(\mathrm{xy}\right)\mathbf{dx}$

Let

$\begin{array}{c}u=\mathrm{xy}\\ \frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}\mathbf{y}\\ \mathbf{dx}\mathbf{=}\frac{1}{y}\mathbf{du}\end{array}$

Now,

$\begin{array}{c}\int \mathbf{sin}\left(\mathrm{xy}\right)\mathbf{dx}\mathbf{=}\frac{1}{y}\int \mathbf{sin}\mathbf{u}\mathbf{du}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{cos}\mathbf{u}}{y}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{y}\end{array}$

Case (2):

$\int \mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{dx}$

Let $\mathbf{u}\mathbf{=}\mathbf{xy}\mathbf{,}\mathbf{dv}\mathbf{=}\mathbf{cos}\left(\mathrm{xy}\right)$.

$\mathbf{du}\mathbf{=}\mathbf{y}\mathbf{dx}\mathbf{,}\mathbf{v}\mathbf{=}\frac{\mathbf{sin}\mathbf{xy}}{y}$

[Note: integrate dv using the method of case (1)]

$\begin{array}{c}\int \mathbf{xy}\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{dx}\mathbf{=}\mathbf{xy}\frac{\mathbf{sin}\mathbf{xy}}{y}\mathbf{-}\int \mathbf{y}\frac{\mathbf{sin}\mathbf{xy}}{y}\mathbf{dx}\\ \mathbf{=}\mathbf{x}\mathbf{sin}\mathbf{xy}\mathbf{+}\frac{\mathbf{cos}\mathbf{xy}}{y}\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Substitute the values in the F.

$\mathbf{F}\mathbf{=}\mathbf{x}\mathbf{sin}\mathbf{xy}\mathbf{+}\frac{\mathbf{cos}\mathbf{xy}}{y}\mathbf{+}\mathbf{g}\left(y\right)$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{-}\frac{\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)}{y}\mathbf{-}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{{\mathbf{y}}^2}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalise the values of N.

$\begin{array}{c}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)\mathbf{-}\frac{\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)}{y}\mathbf{-}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{{\mathbf{y}}^2}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{cos}\left(\mathrm{xy}\right)\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)}{y}\mathbf{+}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{{\mathbf{y}}^2}\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int \left(\mathbf{1}\mathbf{+}\frac{\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)}{y}\mathbf{+}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{{\mathbf{y}}^2}\right)\mathbf{dy}\\ \mathbf{=}\int \mathbf{dy}\mathbf{+}\int \frac{\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)}{y}\mathbf{dy}\mathbf{+}\int \frac{\mathbf{cos}\left(\mathrm{xy}\right)}{{\mathbf{y}}^2}\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}\mathbf{y}\mathbf{-}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{y}\mathbf{+}{\mathbf{C}}_1\end{array}$

Substitute in equation of F.

$\begin{array}{c}\mathbf{F}\mathbf{=}\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{y}\mathbf{+}\mathbf{g}\left(y\right)\\ \mathbf{=}\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{y}\mathbf{+}\mathbf{y}\mathbf{-}\frac{\mathbf{cos}\left(\mathrm{xy}\right)}{y}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{=}\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{C}\end{array}$

So, the solution is$\mathbf{y}\mathbf{+}\mathbf{x}\mathbf{sin}\left(\mathrm{xy}\right)\mathbf{=}\mathbf{C}$