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

$I n p r o b l e m 7 - 16, s o l v e t h e e q u a t i o n . \frac{d x}{d t} - x^{3} = x$

$T h e s o l u t i o n o f t h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \frac{x}{1 + x^{2}} = C e^{t} .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of Separable Differential Equation

$A f i r s t - o r d e r o r d i n a r y d i f f e r e n t i a l e q u a t i o n \frac{d y}{d x} = f (x, y) i s r e f e r r e d t o a s s e p a r a b l e i f t h e f u n c t i o n i n t h e r i g h t - h a n d s i d e o f t h e e q u a t i o n i s e x p r e s s e d a s a p r o d u c t o f t w o f u n c t i o n s g (x) t h a t i s a f u n c t i o n o f x a l o n e a n d h (y) t h a t i s a f u n c t i o n o f y a l o n e . A s e p a r a b l e d i f f e r e n t i a l e q u a t i o n c a n b e e x p r e s s e d a s \frac{d y}{d x} = g (x) h (y) . B y s e p a r a t i n g t h e v a r i a b l e s, t h e e q u a t i o n f o l l o w s \frac{d y}{h (y)} = g (x) d x . T h e n, o n d i r e c t i n t e g r a t i o n o f b o t h s i d e s, t h e s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n i s d e t e r m i n e d .$ $A f i r s t - o r d e r o r d i n a r y d i f f e r e n t i a l e q u a t i o n \frac{d y}{d x} = f (x, y) i s r e f e r r e d t o a s s e p a r a b l e i f t h e f u n c t i o n i n t h e r i g h t - h a n d s i d e o f t h e e q u a t i o n i s e x p r e s s e d a s a p r o d u c t o f t w o f u n c t i o n s g (x) t h a t i s a f u n c t i o n o f x a l o n e a n d h (y) t h a t i s a f u n c t i o n o f y a l o n e . A s e p a r a b l e d i f f e r e n t i a l e q u a t i o n c a n b e e x p r e s s e d a s \frac{d y}{d x} = g (x) h (y) . B y s e p a r a t i n g t h e v a r i a b l e s, t h e e q u a t i o n f o l l o w s \frac{d y}{h (y)} = g (x) d x . T h e n, o n d i r e c t i n t e g r a t i o n o f b o t h s i d e s, t h e s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n i s d e t e r m i n e d .$ uncaught exception: Invalid chunkin file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('f562994a6149b74...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage(' $" width="0" height="0" role="math" style="max-width: none;" localid="1663916781508">$

A first-order ordinary differential equation $\frac{d y}{d x} = f (x, y)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions g(x) that is a function of x alone and h(y) that is a function of y alone.

A separable differential equation can be expressed as $\frac{d y}{d x} = g (x) h (y)$ . By separating the variables, the equation follows $\frac{d y}{h (y)} = g (x) d x$ . Then, on direct integration of both sides, the solution of the differential equation is determined.

Step 2: Solution of the Equation

The given equation is

$\frac{d x}{d t} - x^{3} = x . . . . . . (1)$

After separating the variables, equation (1) can be written as

$\frac{d x}{d t} = x + x^{3} = x (1 + x^{2}) \frac{d x}{x (1 + x^{2})} = d t . . . . . (2)$

Before proceeding for integration, let us decompose the integrand

\frac{1}{x (1 + x^{2})}

into partial fraction decomposition. It follows,

$\frac{1}{x (1 + x^{2})} = \frac{1}{x} - \frac{x}{1 + x^{2}}$

Then, equation (2) becomes,

$(\frac{1}{x} - \frac{x}{1 + x^{2}}) d x = d t . . . . . (3)$

Now, integrate both sides of equation (3). It results,

$\int (\frac{1}{x} - \frac{x}{1 + x^{2}}) d x = \int d t \int \frac{d x}{x} - \int \frac{x d x}{1 + x^{2}} = \int d t \int \frac{d x}{x} - \frac{1}{2} \int \frac{2 x d x}{1 + x^{2}} = \int d t \int \frac{d x}{x} - \int \frac{d (1 + x^{2})}{1 + x^{2}} = \int d t \ln x - \ln (1 + x^{2}) = t + \ln C [\ln C = \ln i n t e g r a t i n g C o n s \tan t] \ln (\frac{x}{1 + x^{2}}) - \ln C = t \ln (\frac{x}{C (1 + x^{2})}) = t \frac{x}{1 + x^{2}} = C e^{t}$ Therefore, the solution of the Therefore, the solution of given equation is $\frac{x}{1 + x^{2}} = C e^{t}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

$I n p r o b l e m 7 - 16, s o l v e t h e e q u a t i o n . \frac{d x}{d t} - x^{3} = x$

Step by Step Solution

Step 1: Concept of Separable Differential Equation

Step 2: Solution of the Equation

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In problem 7-16 , solve the equation. dxdt-x3=x

Step by Step Solution

Step 1: Concept of Separable Differential Equation

Step 2: Solution of the Equation

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$I n p r o b l e m 7 - 16, s o l v e t h e e q u a t i o n . \frac{d x}{d t} - x^{3} = x$