Question

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

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

Short answer

The solution of the given equation is${\mathbf{y}}^2\mathbf{=}{\mathbf{Cx}}^3\mathbf{+}{\mathbf{x}}^2$ and x = 0.

Step-by-step solution

Given information and simplification

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

Evaluate the equation (1).

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

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^2\mathbf{-}\mathbf{3}{\mathbf{y}}^2}{2\mathrm{xy}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Since the given equation is homogeneous. Then, put y = tx and $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\mathbf{t}$.

Substitute the values in equation (2).

role="math" localid="1663948749390" $$\begin{gathered} \begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^2\mathbf{-}\mathbf{3}{\mathbf{y}}^2}{2\mathrm{xy}}\\ \mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\mathbf{t}\mathbf{=}\mathbf{-}\frac{\left(\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{t}}^2\right)}{2t}\\ \mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{\left(\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{t}}^2\right)}{2t}\mathbf{-}\mathbf{t}\end{array} \\ \begin{array}{c}\mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{t}}^2\mathbf{-}\mathbf{2}{\mathbf{t}}^2}{2t}\\ \mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}{\mathbf{t}}^2}{2t}\\ \frac{2t}{{\mathbf{t}}^2\mathbf{-}\mathbf{1}}\mathbf{dt}\mathbf{=}\frac{\mathrm{dx}}{x}\end{array} \end{gathered}$$

Let us assume p = t² - 1. Differentiate with respect to x.

dp = 2t dt .

Substitute the values in equation (1)

$\frac{1}{p}\mathbf{dp}\mathbf{=}\frac{\mathrm{dx}}{x}$

Evaluation method

Now integrate the equation (2) on both sides.

$\begin{array}{c}\int \frac{1}{p}\mathbf{dp}\mathbf{=}\int \frac{\mathrm{dx}}{x}\\ \mathbf{ln}\left|p\right|\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}\\ p=\mathrm{Cx}\end{array}$

Substitute the value of p.

${\mathbf{t}}^2\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{Cx}$

Substitute the value of t.

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

So, the solution is ${\mathbf{y}}^2\mathbf{=}{\mathbf{Cx}}^3\mathbf{+}{\mathbf{x}}^2$