75_pdfsam_math 54 differential equation solutions odd

# 75_pdfsam_math 54 differential equation solutions odd - x 2...

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Exercises 2.4 where c as any nonzero constant. Separating variables, we divided the equation (2.17) by xy .A sar e s u l t ,w e l o s tt w o constant solutions x 0and y 0 (see the discussion on pages 44–45 of Section 2.2 of the text). Thus the orthogonal trajectories for the family 2 x 2 + y 2 = k are x = cy 2 , c 6 =0 , x 0, and y 0. (Note that x 0 can be obtained from x = cy 2 by taking c =0 while y 0 cannot.) (b) First we rewrite the equation de±ning the family of curves in the form F ( x, y )= k by dividing it by x 4 . This yields yx 4 = k . We use (2.17) to set up an equation for the orthogonal trajectories: ∂F ∂x = 4 yx 5 , ∂F ∂y = x 4 x 4 dx ( 4 yx 5 ) dy =0 . Solving this separable equation yields x 4 dx = 4 yx 5 dy =0 xdx = 4 ydy Z xdx = Z ( 4 y ) dy x 2 2 =
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Unformatted text preview: x 2 + 4 y 2 = c . (c) Taking logarithm of both sides of the equation, we obtain ln y = kx ⇒ ln y x = k, and so F ( x, y ) = (ln y ) /x , ∂F/∂x = − (ln y ) /x 2 , ∂F/∂y = 1 / ( xy ). The equation (2.17) becomes 1 xy dx − ± − ln y x 2 ² dy = 0 ⇒ 1 xy dx = − ln y x 2 dy. Separating variables and integrating, we obtain x dx = − y ln y dy ⇒ Z x dx = − Z y ln y dy ⇒ x 2 2 = − y 2 2 ln y + Z y 2 2 · 1 y dy = − y 2 2 ln y + y 2 4 + c 1 ⇒ x 2 2 + y 2 2 ln y − y 2 4 = c 1 ⇒ 2 x 2 + 2 y 2 ln y − y 2 = c, where c := 4 c 1 , and we have used integration by parts to ±nd R y ln y dy . 71...
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