Question

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

$\left({\mathbf{x}}^2\mathbf{-}\mathbf{2}{\mathbf{y}}^{-3}\right)\mathbf{dy}\mathbf{+}\left(\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{3}{\mathbf{x}}^2\right)\mathbf{dx}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given equation is ${\mathbf{x}}^2\mathbf{y}\mathbf{-}{\mathbf{x}}^3\mathbf{+}{\mathbf{y}}^{-2}\mathbf{=}\mathbf{C}$.

Step-by-step solution

Given information and simplification

Given that, $\left({\mathbf{x}}^2\mathbf{-}\mathbf{2}{\mathbf{y}}^{-3}\right)\mathbf{dy}\mathbf{+}\left(\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{3}{\mathbf{x}}^2\right)\mathbf{dx}\mathbf{=}\mathbf{0}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

Then, $\mathbf{M}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{3}{\mathbf{x}}^2\mathbf{,}\mathbf{N}\mathbf{=}{\mathbf{x}}^2\mathbf{-}\mathbf{2}{\mathbf{y}}^{-3}$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{x}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{x}\\ \frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\end{array}$

So, the given equation is exact.

Evaluation method

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{3}{\mathbf{x}}^2$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{3}{\mathbf{x}}^2\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{x}}^2\mathbf{y}\mathbf{-}{\mathbf{x}}^3\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

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

Equalise the values of N.

$\begin{array}{c}{\mathbf{x}}^2\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}{\mathbf{x}}^2\mathbf{-}\mathbf{2}{\mathbf{y}}^{-3}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{y}}^{-3}\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{-}\int \mathbf{2}{\mathbf{y}}^{-3}\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}{\mathbf{y}}^{-2}\mathbf{+}{\mathbf{C}}_1\end{array}$

Substitute in equation of F.

$\begin{array}{c}{\mathbf{x}}^2\mathbf{y}\mathbf{-}{\mathbf{x}}^3\mathbf{+}{\mathbf{y}}^{-2}\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^2\mathbf{y}\mathbf{-}{\mathbf{x}}^3\mathbf{+}{\mathbf{y}}^{-2}\mathbf{=}\mathbf{C}\end{array}$

So, the solution is role="math" localid="1664022924236" ${\mathbf{x}}^2\mathbf{y}\mathbf{-}{\mathbf{x}}^3\mathbf{+}{\mathbf{y}}^{-2}\mathbf{=}\mathbf{C}$