Question

Show that the equation${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$ has no (real-valued) solution.

Short answer

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+{\mathrm{y}}^2+4=0$has no (real-valued) solution.

Step-by-step solution

Simplification of the given differential equation

$\begin{array}{l}{\left(\frac{dy}{dx}\right)}^2=-y^2-4\\ {\left(\frac{dy}{dx}\right)}^2=-\left(y^2+4\right)\\ \frac{dy}{dx}=\sqrt{-\left(y^2+4\right)}\end{array}$

Determining if the given equation has a real-valued solution or not

Now from Step 1, this is clear that the value of $\frac{dy}{dx}$is not real.

Thus ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$ has no (real-valued) solution.