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**Unformatted text preview: **Exercises 2.6
⇒ ⇒ ⇒ µ(t) = exp − 2 dt t = t−2 t−2 u = ⇒ x−2 2 dt = 2 ln |t| + C t = 2t2 ln |t| + Ct2 . 2 d (t−2 u) = ⇒ dt t u = 2t2 ln |t| + Ct2 x ≡ 0 is also a solution, which we lost dividing the equation by x3 . 27. This equation is a Bernoulli equation with n = 2, because it can be written in the form dr 2 − r = r 2 θ −2 . dθ θ Dividing by r 2 and making the substitution u = r −1 , we obtain a linear equation. r −2 dr 2 −1 du 2 ⇒ − − u = θ −2 − r = θ −2 dθ θ dθ θ du 2 2 + u = −θ−2 dθ ⇒ ⇒ µ(θ) = exp dθ θ θ d (θ2 u) = −1 ⇒ θ2 u = −θ + C ⇒ ⇒ dθ θ2 r= C−θ 29. Solving for h and k in the linear system −3h + k − 1 = 0 h+k+3=0 gives h = −1 and k = −2. Thus, we make the substitutions x = u − 1 and y = v − 2, so that dx = du and dy = dv , to obtain (−3u + v ) du + (u + v ) dv = 0. This is the same transformed equation that we encountered in Example 4 on page 77 of the text. There we found that its solution is v 2 + 2uv − 3u2 = C. 85 = θ2 u= −θ + C . θ2 Making back substitution (and adding the lost solution r ≡ 0), we obtain a general solution and r ≡ 0. ...

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