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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 79

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question: In Problems 31-40, solve the initial value problem.
$(2 y^{2} + 4 x^{2}) dx - xydy = 0, y (1) = - 2$

The solution of the given equation is $y = - 2 x \sqrt{2 x^{2} - 1}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information and simplification

Given that, $(2 y^{2} + 4 x^{2}) dx - xy dy = 0, y (1) = - 2 \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Let us check whether the given equation is exact or not.

Then, $M = 2 y^{2} + 4 x^{2}, N = - xy$ .

Differentiate the value of M and N.

$\begin{array}{l} \frac{\partial M}{\partial y} = 4 y \\ \frac{\partial N}{\partial x} = - y \end{array}$

So, the given equation is not exact. Then, find the special integrating factor,

$\begin{matrix} \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{- 4 y - y}{xy} \\ = \frac{y (- 5)}{xy} \\ = - \frac{5}{x} \end{matrix}$

Step 2: Find the exactness

Find the value of $μ (x)$ .

$\begin{matrix} μ (x) = e^{- \int \frac{5}{x} dx} \\ = e^{- 5 \ln |x|} \\ = x^{- 5} \end{matrix}$

Multiply $x^{- 5}$ in equation (1) on both sides.

$\begin{matrix} x^{- 5} (2 y^{2} + 4 x^{2}) dx - {xx}^{- 5} y dy = 0 \\ (2 x^{- 5} y^{2} + 4 x^{- 3}) dx - x^{- 4} y dy = 0 \end{matrix}$

Now again check whether the founded equation is exact or not.

$M = 2 x^{- 5} y^{2} + 4 x^{- 3}, N = x^{- 4} y$

Differentiate the value of M and N.

$\begin{array}{l} \frac{\partial M}{\partial y} = 4 x^{- 5} y \\ \frac{\partial N}{\partial x} = 4 x^{- 5} y \end{array}$

Therefore, the founded equation is exact.

Step 3: Evaluation method

Now, let us assume $M = \frac{\partial F}{\partial x} = 2 x^{- 5} y^{2} + 4 x^{- 3}$ .

Integrate on both sides.

$\begin{matrix} F = \int (2 x^{- 5} y^{2} + 4 x^{- 3}) dx \\ = - 2 x^{- 2} - \frac{y^{2} x^{- 4}}{2} + g (y) \end{matrix}$

Differentiate the F with respect to y.

$\begin{matrix} \frac{\partial F}{\partial y} = - {yx}^{- 4} + g' (y) \\ = N \end{matrix}$

Equalize the values of N.

$\begin{matrix} - {yx}^{- 4} + g' (y) = - {yx}^{- 4} \\ g' (y) = 0 \end{matrix}$

Integrate on both sides.

$\begin{matrix} \int g' (y) = \int 0 dy \\ g (y) = C_{1} \end{matrix}$

Substitute in the equation of F.

$\begin{matrix} - 2 x^{- 2} - \frac{y^{2} x^{- 4}}{2} + C_{1} = 0 \\ - 2 x^{- 2} - \frac{y^{2} x^{- 4}}{2} = C \cdot \cdot \cdot \cdot \cdot \cdot (2) \end{matrix}$

So, the solution is found.

Step 4: Find the initial value

Given that, $y (1) = - 2$ .

Then, x = 1 and y = -2.

Substitute the value in equation (2) to get the value of C.

\begin{matrix} - 2 x^{- 2} - \frac{y^{2} x^{- 4}}{2} = C \\ - 2 {(1)}^{- 2} - \frac{{(- 2)}^{2} {(1)}^{- 4}}{2} = C \\ - 2 - 2 = C \\ C = - 4 \end{matrix}

Substitute the value of C in equation (2).

$\begin{matrix} - 2 x^{- 2} - \frac{y^{2} x^{- 4}}{2} = - 4 \\ - 4 x^{- 2} - y^{2} x^{- 4} = - 8 \\ 4 x^{2} + y^{2} = 8 x^{4} \\ y^{2} = 8 x^{4} - 4 x^{2} \\ y = 4 x^{2} (2 x^{2} - 1) \\ y = \pm \sqrt{4 x^{2} (2 x^{2} - 1)} \\ y = - 2 x \sqrt{2 x^{2} - 1} \end{matrix}$

Since the initial value of y is negative. So that we get the negative value.

So, the solution is $y = - 2 x \sqrt{2 x^{2} - 1}$

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