8_pdfsam_math 54 differential equation solutions odd

# 8_pdfsam_math 54 differential equation solutions odd - , )....

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Chapter 1 which is an identity on ( −∞ , ). Thus φ ( x ) is an explicit solution on ( −∞ , ). (b) We compute dx = d dx ( e x x )= e x 1 . Functions φ ( x )and φ 0 ( x ) are defned ±or all real numbers and dx + φ ( x ) 2 =( e x 1)+( e x x ) 2 =( e x 1)+ ( e 2 x 2 xe x + x 2 ) = e 2 x +(1 2 x ) e x + x 2 1 , which is identically equal to the right-hand side o± the given equation. Thus φ ( x )isan explicit solution on ( −∞ , ). (c) Note that the ±unction φ ( x )= x 2 x 1 is not defned at x = 0. Di²erentiating φ ( x ) twice yields dx = d dx ( x 2 x 1 ) =2 x ( 1) x 2 =2 x + x 2 ; d 2 φ dx 2 = d dx ± dx ² = d dx ( 2 x + x 2 ) =2+( 2) x 3 =2 ( 1 x 3 ) . There±ore x 2 d 2 φ dx 2 = x 2 · 2 ( 1 x 3 ) =2 ( x 2 x 1 ) =2 φ ( x ) , and φ ( x ) is an explicit solution to the di²erential equation x 2 y 0 =2 y on any interval not containing the point x = 0, in particular, on (0
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Unformatted text preview: , ). 3. Since y = sin x + x 2 , we have y = cos x + 2 x and y = sin x + 2. These unctions are defned on ( , ). Substituting these expressions into the dierential equation y + y = x 2 + 2 gives y + y = sin x + 2 + sin x + x 2 = 2 + x 2 = x 2 + 2 or all x in ( , ) . Thereore, y = sin x + x 2 is a solution to the dierential equation on the interval ( , ). 5. Dierentiating x ( t ) = cos 2 t , we get dx dt = d dt (cos 2 t ) = ( sin 2 t )(2) = 2 sin 2 t. 4...
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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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