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I remember reading something about another way to solve an exact differential equation by ignoring particular terms in the equation and then determining if what you have left is the derivative of another function.

Let me illustrate: Suppose we have the first order differential equation $2xy-9x^2+2yy'+x^2y'+y'=0$. If we ignore the following types of terms:

  • Terms only involving $x$
  • Terms involving $y'$ and $x$
  • Scalar terms

Which in this case bring us down to the equation $2xy+x^2y'=0$, then we can check for exactness if what we are left with is the derivative of a function. In this case, $(x^2y)'=2xy+x^2y'=0$, so our equation is exact.

The solution method is also different:

Once we have figured our the function, we can very simply solve the DE:

  1. Re-write the DE like so: $(x^2y)'-9x^2+2yy'+y'=0$
  2. Integrate both sides with respect to $x$: $\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx=0$

$\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx = x^2y-3x^3+y^2+y+c$

And you're done! (This is indeed the correct solution for this differential equation).

I cannot seem to find an example where this method doesn't work, but I am also having trouble understanding WHY it works. Can someone please help? I would also love to hear from whoever I read this from in the first place!

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  • $\begingroup$ Try some of the examples of this search, exact DE with a non-trivial integrating factor, using only your method. $\endgroup$ Commented Jan 15, 2021 at 20:07
  • $\begingroup$ It works because in this case the DE can be written as $\frac{d}{dx}f(x,y)=0$ which is solved by integrating both sides. However this is just a simplification of integrating factor method. $\endgroup$
    – Vasili
    Commented Jan 15, 2021 at 20:23

2 Answers 2

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$$(x^2y)'-9x^2+2yy'+y'=0$$ It works because you have only derivatives: $$(x^2y)'-3(x^3)'+(y^2)'+y'=0$$ $$(x^2y-3x^3+y^2+y)'=0$$ $$x^2y-3x^3+y^2+y=C$$

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If the equation is exact, then that means that there is some function $\phi(x,y)$ such that

$$ \phi(x,y)=c\tag{1} $$

and such that

$$ \frac{\partial\phi}{\partial x}dx+\frac{\partial\phi}{\partial y}dy=0 $$

Alternately,

$$ \frac{\partial\phi}{\partial x}+\frac{\partial\phi}{\partial y}y^\prime=0\tag{2} $$

So we can parse an equation such as

$$ 2xy-9x^2+2yy'+x^2y'+y'=0 $$

into terms containing $y^\prime$ and those not containing $y^\prime$.

$$ 2xy-9x^2 +(2y+x^2+1)y^\prime=0$$

So we know that we are looking for a function $\phi(x,y)$ which simultaneously satisfies

$$ \frac{\partial\phi}{\partial x}=2xy-9x^2 $$ and $$ \frac{\partial\phi}{\partial y}=2y+x^2+1 $$

For the first equation we see that it must be the case that

$$ \phi(x,y)=x^2y-3x^3+c_y $$

where $c_y$ is a 'constant' function of $y$, as far is partial differentiation with respect to $x$ is concerned. Likewise

$$ \phi(x,y)=x^2y+y^2+y+c_x $$

Note that any term of $\phi$ which contains both $x$ and $y$ will occur in both places, even though it occurs only once in the function.

So, using these two clues, we conclude that the solution to equation (1) is

$$ x^2y-3x^3+y^2+y=c$$

and equation (2) is merely a means of parsing the expression to separate out the contributions of the two partials.

Notice that your method is doing the same thing but in a way that makes the process seem somewhat mysterious.

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