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32_pdfsam_math 54 differential equation solutions odd

32_pdfsam_math 54 differential equation solutions odd - C 1...

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Chapter 2 where C 1 is an arbitrary constant and, therefore, C 2 := e C 1 is an arbitrary positive constant. We can rewrite the above solution in the form y = ± C 2 e 2 x cos x = Ce 2 x cos x , (2.1) with C := C 2 or C = C 2 .T h u s C is an arbitrary nonzero constant. The value C =0in (2.1) gives y ( x ) 0, which is, clearly, is also a solution to the diFerential equation. Therefore, the answer to the problem is given by (2.1) with an arbitrary constant C . 11. Separating variables, we obtain dy sec 2 y = dx 1+ x 2 . Using the trigonometric identities sec y =1 / cos y and cos 2 y =(1+cos2 y ) / 2 and integrating, we get dy sec 2 y = dx 1+ x 2 (1 + cos 2 y ) dy 2 = dx 1+ x 2 Z (1 + cos 2 y ) dy 2 = Z dx 1+ x 2 1 2 ± y + 1 2 sin 2 y ² = arctan x + C 1 2 y +sin2 y
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Unformatted text preview: C 1 ⇒ 2 y + sin 2 y = 4 arctan x + C. The last equation de±nes implicit solutions to the given diFerential equation. 13. Writing the given equation in the form dx/dt = x − x 2 , we separate the variables to get dx x − x 2 = dt . Integrate (the left side is integrated by partial fractions, with 1 / ( x − x 2 ) = 1 /x + 1 / (1 − x )) to obtain: ln | x | − ln | 1 − x | = t + c ⇒ ln ³ ³ ³ ³ x 1 − x ³ ³ ³ ³ = t + c ⇒ x 1 − x = ± e t + c = Ce t , where C = e c ⇒ x = Ce t − xCe t ⇒ x + xCe t = Ce t 28...
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