First-Order Exact Differential Equation Shortcut
by Mathandski, May 2, 2025, 9:28 PM
Summary
In this blog post, I will prove an interesting algorithm which serves as a shortcut to solve exact differential equations. This was discovered and introduced to me by my math teacher.
Proposed Shortcut
Given an exact differential equation of the form
![\[M(t, y) + N(t, y) \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/5/a/b/5ab7d1bbf763fbaf19a5acfb75f5a1fc7494d4e6.png)
An algorithm to solve for 
is as follows:
1. Let
and 
where we omit any products of integration.
2. If a certain term appears as an addend in both
and 
, we remove that term from
3. We write
and solve for where 
may be any constant.
Example Problem
![\[2ty - 9t^2 + (2y + t^2 + 1) \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/2/1/d/21d871fb94785dda4ab537e61041dd495e0be58c.png)
Solve for in terms of 
.
Proof
Recall that and after step 
of our algorithm. Our main claim is as follows. We split the terms of into terms exclusively in terms of , terms exclusively in terms of , and other terms dependent on both and . We do the same with .
Remark: For example, for,
![\[F(t, y) = t^2y + 1434 \cos(y^2) + \ln(t + y) + \arctan(y) + t^{2t}\]](//latex.artofproblemsolving.com/8/e/4/8e42229df636b080d176ff934c8ab871eb13d96a.png)
We group 
together, 
together, and 
together.
More formally, for some functions
such that \textbf{each term of 
is dependent on both and }, we may write,
![\[H_1 (t, y) = h_1(t, y) + f_1(t) + g_1(y)\]](//latex.artofproblemsolving.com/7/6/5/765cd275a1991307d99a2c8cf0b68e428ccf640c.png)
![\[H_2 (t, y) = h_2(t, y) + f_2(t) + g_2(y)\]](//latex.artofproblemsolving.com/a/0/0/a00fb96f37e71764b37c81990fed9db21b94d11a.png)
Claim:
![\[h_1(t, y) = h_2(t, y)\]](//latex.artofproblemsolving.com/8/c/d/8cdb892db6fbf474d06537ff633a885b8f25a818.png)
Note that,
![\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]](//latex.artofproblemsolving.com/9/5/6/956e5969c4c7d6d551b0c8bcd8d35c79f1213c1e.png)
Plugging in 
and 
,
![\[\Longrightarrow \frac{\partial}{\partial y} \left( \frac{\partial H_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial H_2}{\partial y} \right) \quad (*)\]](//latex.artofproblemsolving.com/3/d/9/3d9e92c6fc5671f27faa1d192eba5d08bdea5d18.png)
However, notice that,
![\[\frac{\partial H_1}{\partial t} = \frac{\partial}{\partial t} (H_1(t, y)) = \frac{\partial}{\partial t} (h_1(t, y) + g(y)) = \frac{\partial h_1}{\partial t}\]](//latex.artofproblemsolving.com/0/0/c/00c83c8ca9d6875aa6196b03402b37a9b0bb6f24.png)
Analogously,
![\[\frac{\partial H_2}{\partial t} = \frac{\partial h_2}{\partial t}\]](//latex.artofproblemsolving.com/7/6/d/76daefd19b2456d1e0f2db3f5802037a96d92053.png)
Plugging this into 
,
![\[ \frac{\partial}{\partial y} \left( \frac{\partial h_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right)\]](//latex.artofproblemsolving.com/7/f/2/7f2a7c7635b6c7f58fa003da2a28358e5f595f8f.png)
We use Clairaut's Theorem, which states that if
and 
are continuous at 
,
![\[\frac{\partial }{\partial x} \left( \frac{\partial }{\partial y} \left( f(x, y) \right) \right) = \frac{\partial }{\partial y} \left( \frac{\partial }{\partial x} \left( f(x, y) \right) \right)\]](//latex.artofproblemsolving.com/2/9/9/2995196fc0d2629c58257e4a9e4e843c2b5a2c0d.png)
We apply Clairaut's Theorem to the left side of the prior equation,
![\[\iff \frac{\partial}{\partial t} \left( \frac{\partial h_1}{\partial y} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right) \]](//latex.artofproblemsolving.com/f/0/7/f07eba5c82e22ac9c24bef635f7dd645df1ab081.png)
We now do partial integration on both sides. Note that the product of integration becomes an arbitrary function in terms of since we are working with partial derivatives,
![\[\Longrightarrow \frac{\partial h_1}{\partial y} = \frac{\partial h_2}{\partial y} + f(y)\]](//latex.artofproblemsolving.com/1/1/8/1189ce72e64b38a0c2d49a5d3bb6f4806d1edff5.png)
We now integrate with respect to ,
![\[\Longrightarrow h_1(t, y) = h_2(t, y) + \int f(y) \, dy + g(t)\]](//latex.artofproblemsolving.com/c/d/6/cd6c30eaeff568dbbbfd4285a81e1b7b20eb0221.png)
Therefore, 
and 
differ by 
and 
, which are entirely functions in terms of and respectively. However, neither and have any terms which are independent in or from our definition of and . Therefore, the constant functions of integration are both zero and 
as desired. The claim is proved.
Now recall that
and 
are both zero since, under are algorithm, we clarified that we are omitting any constants of integration when writing and . Therefore, we may in fact write,
![\[H_1 (t, y) = h_1(t, y) + f_1(t)\]](//latex.artofproblemsolving.com/a/7/a/a7ae5abc07fbfaaec649fb8a4ea2700a9314005f.png)
![\[H_2 (t, y) = h_2(t, y) + g_2(y)\]](//latex.artofproblemsolving.com/a/0/2/a0289f949b0f273885d1ea07c612c8a3d036475a.png)
We now apply step
of our algorithm, which removes any duplicate terms appearing in both and from .
Claim:
![\[H(t, y) := h_1(t, y) + f_1(t) + g_2(y)\]](//latex.artofproblemsolving.com/b/6/a/b6a7f4395043de9cd158cbf0ea156256dd1bd3d0.png)
Is 
after we apply step of our algorithm.
Indeed, when we added,
Under step, we removed everything in that appeared in . Since 
, the entirety of 
was removed. However, since 
is a function in but there were no terms solely in terms of in , nothing from was removed. As a result, after step ,
![\[H_2(t, y) := g_2(y)\]](//latex.artofproblemsolving.com/3/7/0/37096a723786db65b1b0cc0a2b36317698ed9556.png)
Adding it to
, we get the desired result. The claim is proved.
It now suffices to show that this
has all the solutions to the differential equation.
Claim:
![\[H(t, y) = h_1(t, y) + f_1(t) + g_2(y)\]](//latex.artofproblemsolving.com/6/f/a/6fa6577437e0d43483badedfae65e94342d9d27d.png)
Satisfies 
and 
We may plug it in to prove it. Indeed,
's partial derivative with respect to is 
and adding on a does not change the partial derivative with respect to . The analogous conclusion can be made for 
. The claim is proved.
Therefore, the given differential equation,
![\[\iff \frac{\partial H}{\partial t} + \frac{\partial H}{\partial y} \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/5/a/3/5a3ecc34c1168f3d030d833c577ce03c31d9edf1.png)
Applying the reverse chain rule,
![\[\iff \frac{d}{dt} (H(t, y)) = 0\]](//latex.artofproblemsolving.com/a/b/f/abf0f685c4f34eff460cb3f4971b02680e2f57af.png)
As desired, since is a function of ,
![\[\iff H(t, y) = C\]](//latex.artofproblemsolving.com/6/0/9/60920799b48c87befaf214aa2a3c41c90c0260ec.png)
In this blog post, I will prove an interesting algorithm which serves as a shortcut to solve exact differential equations. This was discovered and introduced to me by my math teacher.
refers to a partial derivative.Proposed Shortcut
Given an exact differential equation of the form
![\[M(t, y) + N(t, y) \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/5/a/b/5ab7d1bbf763fbaf19a5acfb75f5a1fc7494d4e6.png)
An algorithm to solve for 
is as follows:1. Let
and 
where we omit any products of integration.2. If a certain term appears as an addend in both
and 
, we remove that term from 3. We write
and solve for where 
may be any constant.Example Problem
![\[2ty - 9t^2 + (2y + t^2 + 1) \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/2/1/d/21d871fb94785dda4ab537e61041dd495e0be58c.png)
Solve for in terms of 
.![\[\frac{\partial}{\partial t} (2y + t^2 + 1) = 2t = \frac{\partial}{\partial y} (2ty + 9t^2)\]](http://latex.artofproblemsolving.com/8/e/5/8e53a6df127d43062f9123d902a2247f892f9089.png)
Therefore, we may apply the shortcut. We see that, omitting any products of integration,![\[H_1(t, y) = \int 2ty - 9t^2 \, dt = t^2y - 3t^3\]](http://latex.artofproblemsolving.com/3/0/f/30f7a54c2b6c7a3efdc44140a596e1fa6f4a7887.png)
![\[H_2 (t, y) = \int 2y + t^2 + 1 \, dy = y^2 + t^2y + y\]](http://latex.artofproblemsolving.com/a/f/6/af62b6174c443312956e05df52c95f2061d18132.png)
Note that 
appears in both ![\[H_2(t, y) := y^2 + y\]](http://latex.artofproblemsolving.com/f/5/8/f58570302ad0d705e2862d9b383d9e6359bec665.png)
We now write,![\[H_1 (t, y) + H_2(t, y) = C\]](http://latex.artofproblemsolving.com/e/8/e/e8e1b866b4f9ac92669a01fee471b7bcc001c548.png)
![\[\iff y^2 + y + t^2y - 3t^3 = C\]](http://latex.artofproblemsolving.com/a/c/b/acb84a4c2713d5393aea7abe9f5b79b57186c41e.png)
We may now simplify and solve.Proof
Recall that
and after step 
of our algorithm. Our main claim is as follows. We split the terms of into terms exclusively in terms of , terms exclusively in terms of , and other terms dependent on both and . We do the same with .Remark: For example, for,
![\[F(t, y) = t^2y + 1434 \cos(y^2) + \ln(t + y) + \arctan(y) + t^{2t}\]](http://latex.artofproblemsolving.com/8/e/4/8e42229df636b080d176ff934c8ab871eb13d96a.png)
We group 
together, 
together, and 
together.More formally, for some functions
such that \textbf{each term of 
is dependent on both and }, we may write,![\[H_1 (t, y) = h_1(t, y) + f_1(t) + g_1(y)\]](http://latex.artofproblemsolving.com/7/6/5/765cd275a1991307d99a2c8cf0b68e428ccf640c.png)
![\[H_2 (t, y) = h_2(t, y) + f_2(t) + g_2(y)\]](http://latex.artofproblemsolving.com/a/0/0/a00fb96f37e71764b37c81990fed9db21b94d11a.png)
Claim:
![\[h_1(t, y) = h_2(t, y)\]](http://latex.artofproblemsolving.com/8/c/d/8cdb892db6fbf474d06537ff633a885b8f25a818.png)
Note that,
![\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]](http://latex.artofproblemsolving.com/9/5/6/956e5969c4c7d6d551b0c8bcd8d35c79f1213c1e.png)
Plugging in 
and 
,![\[\Longrightarrow \frac{\partial}{\partial y} \left( \frac{\partial H_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial H_2}{\partial y} \right) \quad (*)\]](http://latex.artofproblemsolving.com/3/d/9/3d9e92c6fc5671f27faa1d192eba5d08bdea5d18.png)
However, notice that,![\[\frac{\partial H_1}{\partial t} = \frac{\partial}{\partial t} (H_1(t, y)) = \frac{\partial}{\partial t} (h_1(t, y) + g(y)) = \frac{\partial h_1}{\partial t}\]](http://latex.artofproblemsolving.com/0/0/c/00c83c8ca9d6875aa6196b03402b37a9b0bb6f24.png)
Analogously,![\[\frac{\partial H_2}{\partial t} = \frac{\partial h_2}{\partial t}\]](http://latex.artofproblemsolving.com/7/6/d/76daefd19b2456d1e0f2db3f5802037a96d92053.png)
Plugging this into 
,![\[ \frac{\partial}{\partial y} \left( \frac{\partial h_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right)\]](http://latex.artofproblemsolving.com/7/f/2/7f2a7c7635b6c7f58fa003da2a28358e5f595f8f.png)
We use Clairaut's Theorem, which states that if
and 
are continuous at 
,![\[\frac{\partial }{\partial x} \left( \frac{\partial }{\partial y} \left( f(x, y) \right) \right) = \frac{\partial }{\partial y} \left( \frac{\partial }{\partial x} \left( f(x, y) \right) \right)\]](http://latex.artofproblemsolving.com/2/9/9/2995196fc0d2629c58257e4a9e4e843c2b5a2c0d.png)
We apply Clairaut's Theorem to the left side of the prior equation,
![\[\iff \frac{\partial}{\partial t} \left( \frac{\partial h_1}{\partial y} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right) \]](http://latex.artofproblemsolving.com/f/0/7/f07eba5c82e22ac9c24bef635f7dd645df1ab081.png)
We now do partial integration on both sides. Note that the product of integration becomes an arbitrary function in terms of since we are working with partial derivatives,![\[\Longrightarrow \frac{\partial h_1}{\partial y} = \frac{\partial h_2}{\partial y} + f(y)\]](http://latex.artofproblemsolving.com/1/1/8/1189ce72e64b38a0c2d49a5d3bb6f4806d1edff5.png)
We now integrate with respect to ,![\[\Longrightarrow h_1(t, y) = h_2(t, y) + \int f(y) \, dy + g(t)\]](http://latex.artofproblemsolving.com/c/d/6/cd6c30eaeff568dbbbfd4285a81e1b7b20eb0221.png)
Therefore, 
and 
differ by 
and 
, which are entirely functions in terms of and respectively. However, neither and have any terms which are independent in or from our definition of and . Therefore, the constant functions of integration are both zero and 
as desired. The claim is proved.Now recall that
and 
are both zero since, under are algorithm, we clarified that we are omitting any constants of integration when writing and . Therefore, we may in fact write,![\[H_1 (t, y) = h_1(t, y) + f_1(t)\]](http://latex.artofproblemsolving.com/a/7/a/a7ae5abc07fbfaaec649fb8a4ea2700a9314005f.png)
![\[H_2 (t, y) = h_2(t, y) + g_2(y)\]](http://latex.artofproblemsolving.com/a/0/2/a0289f949b0f273885d1ea07c612c8a3d036475a.png)
We now apply step
of our algorithm, which removes any duplicate terms appearing in both and from .Claim:
![\[H(t, y) := h_1(t, y) + f_1(t) + g_2(y)\]](http://latex.artofproblemsolving.com/b/6/a/b6a7f4395043de9cd158cbf0ea156256dd1bd3d0.png)
Is 
after we apply step of our algorithm.Indeed, when we added,
Under step
, we removed everything in that appeared in . Since 
, the entirety of 
was removed. However, since 
is a function in but there were no terms solely in terms of in , nothing from was removed. As a result, after step ,![\[H_2(t, y) := g_2(y)\]](http://latex.artofproblemsolving.com/3/7/0/37096a723786db65b1b0cc0a2b36317698ed9556.png)
Adding it to
, we get the desired result. The claim is proved.It now suffices to show that this
has all the solutions to the differential equation.Claim:
![\[H(t, y) = h_1(t, y) + f_1(t) + g_2(y)\]](http://latex.artofproblemsolving.com/6/f/a/6fa6577437e0d43483badedfae65e94342d9d27d.png)
Satisfies 
and 
We may plug it in to prove it. Indeed,
's partial derivative with respect to is 
and adding on a does not change the partial derivative with respect to . The analogous conclusion can be made for 
. The claim is proved.Therefore, the given differential equation,
![\[\iff \frac{\partial H}{\partial t} + \frac{\partial H}{\partial y} \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/5/a/3/5a3ecc34c1168f3d030d833c577ce03c31d9edf1.png)
Applying the reverse chain rule,![\[\iff \frac{d}{dt} (H(t, y)) = 0\]](http://latex.artofproblemsolving.com/a/b/f/abf0f685c4f34eff460cb3f4971b02680e2f57af.png)
As desired, since is a function of ,![\[\iff H(t, y) = C\]](http://latex.artofproblemsolving.com/6/0/9/60920799b48c87befaf214aa2a3c41c90c0260ec.png)
instead of 
or 
instead of 
)
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