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

Use the method discussed under “Homogeneous Equations” to solve problems 9- 16.
$(3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy = 0$

Homogeneous equation for the given equation is $\frac{y^{2}}{x^{2}} + \ln |\frac{x^{6}}{y^{2}}| = C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

General form of Homogeneous equation

If the right-hand side of the equation $\frac{dy}{dx} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, $(3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy = 0$

Evaluate it.

$\begin{array}{rcl} (3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy & = & 0 \\ (xy - x^{3} y^{- 1}) dy & = & - (3 x^{2} - y^{2}) dx \\ \frac{dy}{dx} & = & - \frac{3 x^{2} - y^{2}}{xy - x^{3} y^{- 1}} \\ = & \frac{y^{2} - 3 x^{2}}{xy - x^{3} y^{- 1}} \end{array}$

Now take LCM.

$\begin{array}{rcl} \frac{dy}{dx} & = & \frac{y^{3} - 3 {yx}^{2}}{{xy}^{2} - x^{3}} \\ = & \frac{\frac{y^{3}}{x^{3}} - \frac{3 y}{x}}{\frac{y^{2}}{x^{2}} - 1} \\ = & \frac{{(\frac{y}{x})}^{3} - 3 (\frac{y}{x})}{{(\frac{y}{x})}^{2} - 1} \end{array}$

Substitution method

Let us take $v = \frac{y}{x}$

Then $y = v x$

By Differentiating,

role="math" localid="1655179163129" $\begin{array}{rcl} \frac{dy}{dx} & = & v + x \frac{dv}{dx} \\ \frac{v^{3} - 3 v}{v^{2} - 1} & = & v + x \frac{dv}{dx} \\ \frac{v^{3} - 3 v}{v^{2} - 1} - v & = & x \frac{dv}{dx} \\ \frac{v^{3} - 3 v - v^{3} + v}{v^{2} - 1} & = & x \frac{dv}{dx} \end{array}$

$\begin{array}{rcl} x \frac{dv}{dx} & = & - 2 \frac{v}{v^{2} - 1} \\ \frac{v^{2} - 1}{v} dv & = & - \frac{2}{x} dx \\ (v - \frac{1}{v}) dv & = & - \frac{2}{x} dx \end{array}$

Now, integrate on both sides.

$\int (v - \frac{1}{v}) dv = - \int \frac{2}{x} dx \frac{v^{2}}{2} - \ln |v| = - 2 \ln |x| + C v^{2} - 2 \ln |v| = - 4 \ln |x| + 2 C v^{2} - \ln |v^{2}| + \ln |x^{4}| = C$

Substitute $v = \frac{y}{x}$

$\begin{array}{rcl} {(\frac{y}{x})}^{2} - I n | {(\frac{y}{x})}^{2} | + I n | x^{4} | & = & C \\ \frac{y^{2}}{x^{2}} - \ln |\frac{y^{2}}{x^{2}}| + \ln |x^{4}| & = & C \\ \frac{y^{2}}{x^{2}} - \ln |\frac{y^{2}}{x^{2}} \frac{1}{x^{4}}| & = & C \\ \frac{y^{2}}{x^{2}} + \ln |\frac{x^{6}}{y^{2}}| & = & C \end{array}$

Therefore, Homogeneous equation for the given equation is $\frac{y^{2}}{x^{2}} + \ln |\frac{x^{6}}{y^{2}}| = C$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Homogeneous Equations” to solve problems 9- 16.
$(3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy = 0$

Step by Step Solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Homogeneous Equations” to solve problems 9- 16. 3x2-y2 dx+xy-x3y-1dy=0

Step by Step Solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

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Use the method discussed under “Homogeneous Equations” to solve problems 9- 16.
$(3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy = 0$