Lets have y : R → R 2 and that f : R 2 → R 2 , and lets assume that f ( y ( x ) ) is given and that y ( x ) = y ( x 0 ) + ∫ x 0 x f ( y ( t ) ) d tI'm a bit confused how there can be a function of y ( t ) inside of the function definition for y ( x ).I took the example that y ( x ) = ( x 2 , x ) and f ( y , z ) = ( y + z , y − z ) ⇒ f ( y ( x ) ) = f ( x 2 , x ) = ( x 2 + x , x 2 − x )And now if we follow the definition of y ( x ) we get: y ( x ) = y ( x 0 ) + ∫ x 0 x f ( y ( t ) ) d t y ( x ) = y ( x 0 ) + ∫ x 0 x ( t 2 + t , t 2 − t ) d t ⇒ y ( x ) = y ( x 0 ) + ( 1 3 t 3 + 1 2 t 2 , 1 3 t 3 − 1 2 t 2 ) | x 0 x ⇒ y ( x ) = ( 1 3 x 3 + 1 2 x 2 + C 1 , 1 3 x 3 − 1 2 x 2 + C 2 )Where is mistake?

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Lets have   y  :      R    →            R        2   and that   f  :            R        2    →            R        2  , and lets assume that       f    (    y    (    x    )    )   is given and that   y  (  x  )  =  y  (      x    0    )  +      ∫                  x        0              x    f  (  y  (  t  )  )  d  tI&#039;m a bit confused how there can be a function of   y  (  t  ) inside of the function definition for   y  (  x  ).I took the example that   y  (  x  )  =  (      x    2    ,  x  ) and   f  (  y  ,  z  )  =  (  y  +  z  ,  y  −  z  )  ⇒  f  (  y  (  x  )  )  =  f  (      x    2    ,  x  )  =  (      x    2    +  x  ,      x    2    −  x  )And now if we follow the definition of   y  (  x  ) we get:  y  (  x  )  =  y  (      x    0    )  +      ∫                  x        0              x    f  (  y  (  t  )  )  d  t  y  (  x  )  =  y  (      x    0    )  +      ∫                  x        0              x    (      t    2    +  t  ,      t    2    −  t  )  d  t  ⇒  y  (  x  )  =  y  (      x    0    )  +  (      1    3        t    3    +      1    2        t    2    ,      1    3        t    3    −      1    2        t    2    )                    |                            x        0              x    ⇒  y  (  x  )  =  (      1    3        x    3    +      1    2        x    2    +      C    1    ,      1    3        x    3    −      1    2        x    2    +      C    2    )Where is mistake?

Solomon Fernandez · Accepted Answer

simplest example is to fix   f  (  x  ,  y  )  =  (  x  +  1  ,  y  +  1  ) and let   y  (  x  )  =  (      e    x    −  1  ,      e    x    −  1  ).

Lets have y:R->R^2 and that f:R^2→R^2, and lets assume that f(y(x)) is given and that y(x)=y(x0)+∫^x_(x_0) f(y(t))dt

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