The question asks to find y given y(x) is a differentiable function satisfying: d y d x =- 2 x y 4 , y(0)= 1 3 and y ( x ) &gt; 0 , and explain why it is unique.I assume the steps are to integrate d y d x to get a function for y(x), then use y ( 0 ) = 1 / 3 to find the constant and then set y ( x ) &gt; 0 to find y but i keep getting stuck.A differential equation solution gives y(x)= 1 c 1 + 3 x 2 3 , which I can use y ( 0 ) to find c 1 but then I'm not sure how to get y from that since the equation doesn't have any y's in it.A similar question I found on this site showed to rearrange and integrate each side like ∫ d y y 4 = ∫ − 2 x d x which gives − 1 3 y 3 + c 1 = − x 2 + c 2 , but then I don't know how to get y ( x ) from that frunction, it seems further from the solution than my first try.

Question

The question asks to find y given y(x) is a differentiable function satisfying:            d      y              d      x      =-  2  x      y    4   , y(0)=      1    3   and   y  (  x  )  &amp;gt;  0 , and explain why it is unique.I assume the steps are to integrate             d      y              d      x       to get a function for y(x), then use   y  (  0  )  =  1      /    3 to find the constant and then set   y  (  x  )  &amp;gt;  0 to find y but i keep getting stuck.A differential equation solution gives y(x)=      1                            c          1                +        3                  x          2                    3       , which I can use   y  (  0  ) to find       c    1   but then I&#039;m not sure how to get y from that since the equation doesn&#039;t have any y&#039;s in it.A similar question I found on this site showed to rearrange and integrate each side like   ∫            d      y              y      4        =  ∫  −  2  x  d  x which gives   −      1          3              y        3              +      c    1  =  −      x    2    +      c    2   , but then I don&#039;t know how to get   y  (  x  ) from that frunction, it seems further from the solution than my first try.

Kaylie Mcdonald · Accepted Answer

That&#039;s just a simple manipulation:                    −                  1                      3                          y              3                                      +                  c          1                =        −                  x          2                +                  c          2                                            ⟺                −                  1                      3                          y              3                                      =        −                  x          2                +                  c          2                −                  c          1                                                          ⟺                          1                      3                          y              3                                      =                  x          2                +                  c          1                −                  c          2                                                          ⟺                          y          3                =                  1                      3                          x              2                        +            3            (                          c              1                        −                          c              2                        )                                                                    ⟺                y        =                  1                                    3                              x                2                            +              3              (                              c                1                            −                              c                2                            )                        3                          .            Now it&#039;s just note that   3  (      c    1    −      c    2    ) is constant, say   k, and we obtain  y  =      1                  3                  x          2                +        k            3        .