104_pdfsam_math 54 differential equation solutions odd

104_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 2 ⇒ F (x, y ) = − x−4 y 2 − 2x−2 = C. 2 We find C by substituting the initial condition, y (1) = −2: − So, the solution is − x−4 y 2 − 2x−2 = −4 2 ⇒ y 2 + 4x2 = 8x4 y 2 = 8x4 − 4x2 = 4x2 2x2 − 1 √ y = −2x 2x2 − 1 , (1)−4 (−2)2 − 2(1)−2 = C 2 ⇒ C = −4 . ⇒ ⇒ where, taking the square root, we have chosen the negative sign because of the initial negative value for y . 37. In this equation with linear coefficients we make a substitution x = u + h, y = v + k with h and k such that 2h − k = 0 h+k = 3 Therefore, (2u − v )du + (u + v )dv = 0 v − 2u (v/u) − 2 dv = = ⇒ du v+u (v/u) + 1 ⇒ z = v/u, v = uz, v = z + uz z−2 dz z2 + 2 dz = ⇒ u =− ⇒ z+u du z+1 du z+1 z+1 du ⇒ dz = − . z2 + 2 u Integration yields z+1 dz = − z2 + 2 100 du u ⇒ z dz + z2 + 2 dz =− z2 + 2 du u ⇒ k = 2h h + (2h) = 3 ⇒ k = 2, h = 1. ...
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