47_pdfsam_math 54 differential equation solutions odd

47_pdfsam_math 54 differential equation solutions odd - y...

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Exercises 2.3 11. Choosing t as the independent variable and y as the dependent variable, we put the equation put into standard form: t + y +1 dy dt =0 dy dt y = t +1 . (2.8) Thus P ( t ) ≡− 1andso µ ( t )=exp ±R ( 1) dt ² = e t . We multiply both sides of the second equation in (2.8) by µ ( t ) and integrate. This yields e t dy dt e t y =( t +1) e t d dt ( e t y ) =( t +1) e t e t y = Z ( t +1) e t dt = ( t +1) e t + Z e t dt = ( t +1) e t e t + C = ( t +2) e t + C y = e t ( ( t +2) e t + C ) = t 2+ Ce t , where we have used integration by parts to Fnd R ( t +1) e t dt . 13. In this problem, the independent variable is y and the dependent variable is x . So, we divide the equation by y to rewrite it in standard form.
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Unformatted text preview: y dx dy + 2 x = 5 y 2 ⇒ dx dy + 2 y x = 5 y 2 . Therefore, P ( y ) = 2 /y and the integrating factor, µ ( y ), is µ ( y ) = exp ³Z 2 y dy ´ = exp (2 ln | y | ) = | y | 2 = y 2 . Multiplying the equation (in standard form) by y 2 and integrating yield y 2 dx dy + 2 y x = 5 y 4 ⇒ d dy ( y 2 x ) = 5 y 4 ⇒ y 2 x = Z 5 y 4 dy = y 5 + C ⇒ x = y − 2 ( y 5 + C ) = y 3 + Cy − 2 . 15. To put this linear equation in standard form, we divide by ( x 2 + 1) to obtain dy dx + x x 2 + 1 y = x x 2 + 1 . (2.9) 43...
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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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