Trivial ODE and trivial question: x ( y + 4 ) + d y d x = 0with initial conditions y = − 5 , x = 0After we separate variables we get: − d y y + 4 = x d xIntegrate left and write parts: − ln ⁡ ( y + 4 ) = 1 2 x 2 + CHere we see that if y = − 5 we have a log of negative number. Trying not to think about it, I proceed as follows:Exponentiation of both parts: y = e − 1 / 2 x 2 − C − 4And then: y = e − 1 / 2 x 2 e − C − 4 y = e − 1 / 2 x 2 C − 4(this new C to e − C and cannot be negative) − 5 = e 0 C − 4 C = − 1(but we see that it is in fact negative under given initial conditions)which gives us the correct answer: y = − e − 1 / 2 x 2 − 4Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative C (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?

Question

Trivial ODE and trivial question:     x  (  y  +  4  )  +            d      y              d      x        =  0with initial conditions   y  =  −  5  ,  x  =  0After we separate variables we get:      −            d      y              y      +      4        =  x    d  xIntegrate left and write parts:  −  ln  ⁡  (  y  +  4  )  =      1    2        x    2    +  CHere we see that if   y  =  −  5 we have a log of negative number. Trying not to think about it, I proceed as follows:Exponentiation of both parts:  y  =      e          −      1          /      2              x        2            −      C        −  4And then:  y  =      e          −      1          /      2              x        2                  e          −      C        −  4  y  =      e          −      1          /      2              x        2              C  −  4(this new   C to       e          −      C       and cannot be negative)  −  5  =      e      0    C  −  4  C  =  −  1(but we see that it is in fact negative under given initial conditions)which gives us the correct answer:  y  =  −      e          −      1          /      2              x        2              −  4Now I feel that I&#039;ve cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative C (which shouldn&#039;t be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?

Mya Hurst · Accepted Answer

Re: your first integration: remember that   ∫                    f        ′            (      y      )              f      (      y      )        d  y  =  ln  ⁡  (  |  f  (  y  )  |  )  +  CSo in your case, the LHS should result in   −  ln  ⁡  (  |  y  +  4  |  )  .

Trivial ODE and trivial question: <mtext> </mtext> x ( y + 4 ) +

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