Exact Differential EquationsI was revising differential equations and came across the topic of exact differential equations. I have a doubt concerning it. Suppose the differential equation M ( x , y ) d x + N ( x , y ) d y = 0 is exact. Then the solution is given by: ∫ M d x + ∫ ( N − ∂ ∂ y ∫ M d x ) d y = c. I understand that the integrand in the second term is a function of y alone and also understand the derivation of this solution. What I don't understand is the following paragraph:My book then says "Since all the terms of the solution that contain x must appear in ∫ M d x, its derivative w.r.t. y must have all the terms of N that contain x. Hence the general rule to be followed is: Integrate ∫ M d x as if y were constant. Also integrate the terms of N that do not contain x w.r.t. y. Equate the sum of these integrals to a constant."

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Exact Differential EquationsI was revising differential equations and came across the topic of exact differential equations. I have a doubt concerning it. Suppose the differential equation   M  (  x  ,  y  )  d  x  +  N  (  x  ,  y  )  d  y  =  0 is exact. Then the solution is given by:   ∫  M  d  x  +  ∫  (  N  −      ∂          ∂      y        ∫  M  d  x  )  d  y  =  c. I understand that the integrand in the second term is a function of y alone and also understand the derivation of this solution. What I don&#039;t understand is the following paragraph:My book then says &quot;Since all the terms of the solution that contain x must appear in   ∫  M  d  x, its derivative w.r.t. y must have all the terms of N that contain x. Hence the general rule to be followed is: Integrate   ∫  M  d  x as if y were constant. Also integrate the terms of N that do not contain x w.r.t. y. Equate the sum of these integrals to a constant.&quot;

benyaep17 · Accepted Answer

Step 1I believe that with exact differential equation you mean that there is some function U such that  d  U  (  x  ,  y  )  =  M  (  x  ,  y  )  d  x  +  N  (  x  ,  y  )  d  y  ,,or e.g. the case when       M    y    =      N    x   in the simply connected domain of             R        2  , which is sufficient for U to exist. Then all solutions clearly are described as   U  (  x  ,  y  )  =  c and the only question is how to find U.Step 2Using the theory of curve integrals,  U  (  x  ,  y  )  =  U  (      x    0    ,      y    0    )  +      ∫    γ    M  (  x  ,  y  )  d  x  +  N  (  x  ,  y  )  d  ywhere   γ is any (piecewise-smooth) path from   (      x    0    ,      y    0    ) to (x,y). The most simple one is to take   γ  =  A  B  ∪  B  C where   A  =  (      x    0    ,      y    0    )  ,  B  =  (  x  ,      y    0    ) and   C  =  (  x  ,  y  ). Then your integral is decomposed into two integrals since   d  x            |              B      C        =  0 and   d  y            |              A      B        =  0. Then the formula you wrote just follows from the theory of curve integrals.

JetssheetaDumcb · Accepted Answer

Step 1There seemed to be a misunderstanding as people tried to explain to me why   ∫  M  d  x  +  ∫  (  N  −      ∂          ∂      y        ∫  M  d  x  )  d  y  =  c is the solution of the exact ODE, something which I had already understood perfectly. My problem was with the next statement in the book which gave a working rule that essentially said that the solution could be expressed as   ∫  M  d  x (y constant)   +  ∫      N    ′    d  y  =  c (N&#039; are the terms of N not containing x) is the solution. After some online searches I have hence discovered the solution. The book is wrong.Step 2The rule it quotes works so often in practice that people adopt it but there are cases when it fails and we have to take recourse to   ∫  M  d  x  +  ∫  (  N  −      ∂          ∂      y        ∫  M  d  x  )  d  y  =  c to write the solution. For example the ODE             d      x                      x        2            +              y        2              +  (      1    y    −      x          y                        x          2                +                  y          2                      )  d  y  =  0 would give a wrong answer on applying the working rule and we have to take recourse to directly computing   ∫  M  d  x  +  ∫  (  N  −      ∂          ∂      y        ∫  M  d  x  )  d  y  =  c.

Exact Differential Equations. I was revising differential equations and came across the topic of exact differential equations. I have a doubt concerning it. Suppose the differential equation M(x,y)dx+N(x,y)dy=0 is exact.

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