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Example solving non-linear first order ODE

Nasser M. Abbasi

June 16, 2017 compiled on — Friday June 16, 2017 at 03:46 PM
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Write as

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Where

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Check if exact

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Since ∂M-(t,y) ⁄= ∂N(t,y)
  ∂y       ∂t  then Not exact. Trying integrating factor A = ∂N∂t − ∂∂My-=-− 32y12
       M     y32− a32   , Since it is a function of y  alone, then it (1) can be made exact. The integrating factor is

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Multiplying (1) by this integrating factor, now it becomes exact

μM (t,y) dt+ μN (t,y)dy = 0

Now we follow standard method for solving exact ODE. Let

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From (2)

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Substituting this into (3) to solve for f (y)

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Hence the solution from (4) is

           √ -      (     ∘ y)
U = − t+ − 2√-3arctan  1-+√2--a  − -2√--ln(√a-− √y-)+ -1√--ln(a+ √ay-+ y)+ C
         3  a            3       3  a              3 a

But ddUt = 0  , hence U = C1  . Therefore, collecting constants into one, the solution is (implicit form)

    √ -      (     ∘ -)
   2--3-       1+-2--ya-    -2--  (√ -- √ -)  --1--     √ ---
t+ 3√ a arctan   √ 3    +  3√ a ln   a−   y − 3√ a ln(a + ay+ y) = C

From initial conditions

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Hence final solution for y(t)  in implicit form is

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