Question

\(x y^{2}=1\) \(y^{2}+2 x y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-y}{2 x}\) \(-\frac{d x}{d y}=\frac{2 x}{y}=\frac{2}{y^{3}}\) \(y-y_{1}=\frac{2}{y_{1}^{3}}\left(x-x_{1}\right)\) \(+y_{1}^{4}=2 x_{1}\) \(y_{1}^{6}=2\) \(y_{1}=\pm 2^{1 / 6}\) \(x_{1}=\pm 2^{-\sqrt{3}}\)

Short answer

2) What value of \(y_1\) did we derive from the calculations? 3) What value of \(x_1\) did we derive from the calculations?

Step-by-step solution

Identify the equations given

The equations given are: 1) \(xy^2 = 1\) 2) \(y^2 + 2xy \frac{dy}{dx} = 0\) 3) \(\frac{dy}{dx} = \frac{-y}{2x}\) 4) \(-\frac{dx}{dy} = \frac{2x}{y} = \frac{2}{y^3}\) 5) \(y - y_1 = \frac{2}{y_1^3}(x - x_1)\) 6) \(y_1^4 = 2x_1\) 7) \(y_1^6 = 2\)

Find the value of \(y_1\)

We will use equation (7) to evaluate the value of \(y_1\). $$y_1^6 = 2$$ Now, we can find the value of \(y_1\) using the equation above: $$y_1 = \pm 2^{\frac{1}{6}}$$

Find the value of \(x_1\)

Now that we have the value of \(y_1\), let's use equation (6) to find the value of \(x_1\). $$y_1^4 = 2x_1$$ Substitute \(y_1\) with \(\pm 2^{\frac{1}{6}}\): $$(\pm 2^{\frac{1}{6}})^4 = 2x_1$$ This gives: $$x_1 = \pm 2^{-\sqrt{3}}$$

Conclusion

Using the above steps, we have determined the related general solution of the given system of equations. We found \(y_1 = \pm 2^{\frac{1}{6}}\) and \(x_1 = \pm 2^{-\sqrt{3}}\).