I want to solve the differential equation $\frac{dy}{dx}=y$, but I only know the informal way of doing it by splitting the differential.

$$\frac{dy}{dx}=y$$ $$\frac{1}{y}dy=dx$$ $$\int\frac{1}{y}dy=\int dx$$ $$\ln|y|=x+C$$ $$y=Ce^x$$

I know that this is the correct answer because exponential functions are their own derivatives. However, this is not the correct way to do it, because $\frac{dy}{dx}$ is not a fraction, but an operator on $y$ instead. When I try to instead try to take the integral with respect to $x$, I get

$$\frac{dy}{dx}=y$$ $$\frac{dy}{dx}dx=ydx$$ $$\int\frac{d}{dx}(y)dx=\int ydx$$ $$y=yx+C$$

I know that this is not the correct answer, because taking the derivative does not give back the original differential equation. Which step is incorrect? Is it the way that I am setting up the integral, my integration of the derivative of $y$, or is it integrating $y$ with respect to $x$?

If the last option of my integration of $y$ on the right side being incorrect, why is that? Shouldn't this integral with respect to $x$ "consider" $y$ to be a constant?

How is this done the formal way?


4 Answers 4


You are doing the below manipulation: $$ \int ydx = yx+C $$ This is not correct because $y$ is not a constant with respect to $x$, $y$ is a function of $x$. For example, let's say $y=x^2$. Then, the above equation implies: $$ \int x^2 dx = x^2\cdot x+C=x^3+C $$ which is clearly not correct, since the antiderivative of $x^2$ is $x^3/3+C$.

Ultimately, the reason "splitting the differential" works is because of $u$-substitution. I think showing how $u$-substitution applies in this situation more explicitly will clarify: $$ \begin{align*} & \frac{dy}{dx}=y \\ \implies & \frac{1}{y}\frac{dy}{dx}=1 \\ \implies & \int \frac{1}{y}\frac{dy}{dx}dx=\int dx \end{align*} $$ On the left-hand side, we have $\frac{dy}{dx}dx$. By $u$-substitution, we can replace this with just $dy$, so we get: $$ \int \frac{1}{y}dy=\int dx $$ and this is exactly the equation we would have ended up with if we had split the differential before taking the integral with respect to $dx$.

Whenever we "split the differential," we are essentially using $dy$ as a shorthand for $\frac{dy}{dx}dx$, and then when we integrate, replacing $dy$ with $\frac{dy}{dx}dx$ is justified because of $u$-substitution, which is why the approach of splitting the differential has a rigorous basis.

  • $\begingroup$ Thank you, this really explains how this works! Is the chain rule applicable for higher order differential equations? My Calculus course hasn't gotten to higher order differential equations, so I am wondering how the chain rule applies when $$\frac{d^2y}{dx^2}$$ is involved. $\endgroup$
    – Alteria
    Commented Apr 3 at 5:33
  • $\begingroup$ @Alteria Solving higher-order differential equations can be very tricky and I've only solved higher-order differential equations where the coefficients on the $dy/dx, d^2y/dx^2$, etc. were constant, so I didn't need to separate the differential. However, the same principle of chain rule applies: If we let $u=dy/dx$, then for example, we could apply chain rule in this manner: $$ \frac{du}{dx}=\frac{d^2y}{dx^2}\implies du=\frac{d^2y}{dx^2}dx $$ This is also a good post to read: math.stackexchange.com/a/4602231/307483 It applies chain rule in a higher-order DE to the fraction $du/dy$. $\endgroup$ Commented Apr 3 at 5:44

The question has already been answered by @NobleMushtak. Just to put it there, here is another "right" way to write these arguments.

Step 1: Use the informal calculation to find the solution, in this case $y=Ce^x$.

Step 2: Express the constant in terms of everything else, and call it a new function. In this case, define $\phi(x)=ye^{-x}$.

Step 3: Differentiate the function, it must be constant, and so you are done. In this case, $$\phi^\prime(x)=y^\prime e^{-x}-ye^{-x}=(y^\prime-y)e^{-x}=0$$ and so $\phi(x)=C$, or $y=Ce^x$.


Try this:






$y=e^{x+C}=A\times e^{x}$

Does that work better?

The $\frac{dy}{dx}\times\frac{dx}{dy}=1$ I believe is justified by the chain rule.

  • $\begingroup$ This does answer my question, thank you! But what do you mean that dy/dx * dx/dy is justified by the chain rule? Does this approach work for other differential equations as well? $\endgroup$
    – Alteria
    Commented Apr 3 at 5:25
  • 1
    $\begingroup$ @Alteria Chain rule for derivatives says that: $$\frac{du}{dv}\cdot \frac{dv}{dw}=\frac{du}{dw}$$ If we use $u=y$, $v=x$, and $w=y$, then we get $$\frac{dy}{dx}\cdot \frac{dx}{dy}=\frac{dx}{dx}=1$$ $\endgroup$ Commented Apr 3 at 5:26
  • $\begingroup$ The chain rule uses limits to prove it is a valid rule for well-behaved functions but once proved, it says (to a non-Mathematician) that multiplication of derivatives behaves a lot like the multiplication of fractions. Hence the product of the two derivatives should be equal to 1. $\endgroup$
    – Red Five
    Commented Apr 3 at 5:28
  • 1
    $\begingroup$ Right, I had forgotten about that. I'm only in my Calculus 1 course and I only remember the chain rule being used for composed functions. i.e the derivative of f(g(x)) being f'(g(x))g'(x) $\endgroup$
    – Alteria
    Commented Apr 3 at 5:28
  • $\begingroup$ How does $\frac{\mathrm dx}{\mathrm dy}=\frac{1}{y}$ implies that $x+C=\ln y$ as in your 4th and 5th line ? Isn't that exactly the question OP asked ? $\endgroup$ Commented Apr 3 at 15:59

$y' = y$ $\forall x \in \mathbb{R}$
$y'e^x=ye^x$ $\forall x \in \mathbb{R}$
$\frac{y'e^x-ye^x}{e^{2x}}=0$ $\forall x \in \mathbb{R}$
notice that $\frac{y'e^x-ye^x}{e^{2x}}$ is just the derivative of $\frac{y}{e^{x}}$ so:
$\frac{y}{e^{x}}+c_1=c_2$ for some reals $c_1$ and $c_2$ (notice that $e^x$ is non-zero for any real)
$y=ce^x$ for some real $c=c_2-c_1$

notice that every line is equivalent to its predecessor. Rigorously solving these is just about introducing the exp() function in the right way and avoiding the division over a function that is not defined to be non-zero, imo

  • 1
    $\begingroup$ Where did the denominator for $\frac{y'e^x-ye^x}{e^{2x}}=0$ come from? The step before this should've given $y'e^x-ye^x=0$, right? $\endgroup$
    – Alteria
    Commented Apr 4 at 1:56
  • $\begingroup$ @Alteria yeah substract $ye^x$ from both sides then divide both by $e^{2x}$ which is a non zero quantity for any real $x$. 0 divided stays 0 any way, $\endgroup$
    – zaknenou
    Commented Apr 4 at 20:35

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