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Q19 E

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Found in: Page 14

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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.

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

See the step by step solution

## Step 1: Simplification of the given differential equation

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

## Step 2: 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.

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