Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Simplification of the given differential equation

$\begin{array}{l} {(\frac{d y}{d x})}^{2} = - y^{2} - 4 \\ {(\frac{d y}{d x})}^{2} = - (y^{2} + 4) \\ \frac{d y}{d x} = \sqrt{- (y^{2} + 4)} \end{array}$

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 $\frac{d y}{d x}$ is not real.

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

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

Step by Step Solution

Step 1: Simplification of the given differential equation

Step 2: Determining if the given equation has a real-valued solution or not

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Show that the equation (dydx)2+y2+4=0 has no (real-valued) solution.

Step by Step Solution

Step 1: Simplification of the given differential equation

Step 2: Determining if the given equation has a real-valued solution or not

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