10_pdfsam_math 54 differential equation solutions odd

10_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 1 Therefore, the function y (x) defined by exy + y = x − 1 is an implicit solution to the given differential equation. 13. Differentiating the equation sin y + xy − x3 = 2 implicitly with respect to x, we obtain y cos y + xy + y − 3x2 = 0 ⇒ (cos y + x)y = 3x2 − y ⇒ y= 3x2 − y . cos y + x Differentiating the second equation above again, we obtain (−y sin y + 1)y + (cos y + x)y = 6x − y ⇒ ⇒ (cos y + x)y = 6x − y + (y )2 sin y − y = 6x − 2y + (y )2 sin y 6x − 2y + (y )2 sin y . y= cos y + x 3x2 − y , cos y + x Multiplying the right-hand side of this last equation by y /y = 1 and using the fact that y= we get y 6x − 2y + (y )2 sin y y · 2 − y )/(cos y + x) cos y + x (3 x 2 3 6xy − 2(y ) + (y ) sin y . = 3x2 − y = Thus y is an implicit solution to the differential equation. 15. We differentiate φ(x) and substitute φ and φ into the differential equation for y and y . This yields φ(x) = Ce3x + 1 ⇒ dφ(x) = C e3x + 1 = 3Ce3x ; dx dφ − 3φ = 3Ce3x − 3 C e3x + 1 = (3C − 3C )e3x − 3 = −3, dx which holds for any constant C and any x on (−∞, ∞). Therefore, φ(x) = Ce3x + 1 is a one-parameter family of solutions to y − 3y = −3 on (−∞, ∞). Graphs of these functions for C = 0, ±0.5, ±1, and ±2 are sketched in Figure 1-A. 6 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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