Question

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

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

Short answer

The solution of the given equation is ${\left({\mathbf{x}}^2\mathbf{+}\mathbf{4}{\mathbf{y}}^2\right)}^3{\mathbf{x}}^2\mathbf{=}\mathbf{C}$

Step-by-step solution

Given information and simplification

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

Evaluate the equation (1).

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

Let us take $\mathbf{y}\mathbf{=}\mathbf{tx}$. Differentiate with respect to x.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\mathbf{t}$

Then,

$\begin{array}{c}\mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\mathbf{t}\mathbf{=}\mathbf{-}\frac{\left(\mathbf{1}\mathbf{+}{\mathbf{t}}^2\right)}{3t}\\ \mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{\left(\mathbf{1}\mathbf{+}{\mathbf{t}}^2\right)}{3t}\mathbf{-}\mathbf{t}\\ \mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{-}\left(\mathbf{1}\mathbf{+}{\mathbf{t}}^2\right)\mathbf{-}\mathbf{3}{\mathbf{t}}^2}{3t}\end{array}$

$\begin{array}{c}\mathbf{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{t}}^2}{3t}\\ \frac{3t}{\mathbf{1}\mathbf{+}\mathbf{4}{\mathbf{t}}^2}\mathbf{dt}\mathbf{=}\mathbf{-}\frac{\mathrm{dx}}{x}\end{array}$

Let us take$\mathbf{p}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{4}{\mathbf{t}}^2$ and $\mathbf{dp}\mathbf{=}\mathbf{8}\mathbf{t}\mathbf{dt}$.

$\frac{3}{8p}\mathbf{dp}\mathbf{=}\mathbf{-}\frac{\mathrm{dx}}{x}$

Now integrate on both sides.

$\begin{array}{c}\int \frac{3}{8p}\mathbf{dp}\mathbf{=}\mathbf{-}\int \frac{\mathrm{dx}}{x}\\ \frac{3}{8}\mathbf{ln}\left|p\right|\mathbf{=}\mathbf{-}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}\\ \mathbf{p}\mathbf{=}\mathbf{C}\mathbf{\text{'}}{\mathbf{x}}^{\mathbf{-}\frac{8}{3}}\end{array}$

Evaluation method

Substitute the value of p.

$\begin{array}{c}\mathbf{1}\mathbf{+}\mathbf{4}{\mathbf{t}}^2\mathbf{=}\mathbf{C}\mathbf{\text{'}}{\mathbf{x}}^{\mathbf{-}\frac{8}{3}}\\ {\left(\mathbf{1}\mathbf{+}\mathbf{4}{\mathbf{t}}^2\right)}^3\mathbf{=}\mathbf{C}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{x}}^{-8}\end{array}$

Substitute the value of t.

$\begin{array}{c}{\left(\frac{{\mathbf{x}}^2\mathbf{+}\mathbf{4}{\mathbf{y}}^2}{{\mathbf{x}}^2}\right)}^3\mathbf{=}\mathbf{C}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{x}}^{-8}\\ {\left({\mathbf{x}}^2\mathbf{+}\mathbf{4}{\mathbf{y}}^2\right)}^3{\mathbf{x}}^2\mathbf{=}\mathbf{C}\end{array}$

So, the solution is ${\left(x^2+4y^2\right)}^3{\mathbf{x}}^2\mathbf{=}\mathbf{C}$