# How do you actually solve a linear homogeneous differential equation rigorously without splitting the differential

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?

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.

• 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. Apr 3 at 5:33
• @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 $du/dy$. 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:

$$\frac{dy}{dx}=y$$

$$\frac{dy}{dx}\times\frac{dx}{dy}=y\times\frac{dx}{dy}$$

$$1=y\times\frac{dx}{dy}$$

$$\frac{dx}{dy}=\frac{1}{y}$$

$$x+C=\log_{e}{y}$$

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

Does that work better?

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

• 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? Apr 3 at 5:25
• @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 $$\frac{dy}{dx}\cdot \frac{dx}{dy}=\frac{dx}{dx}=1$$ Apr 3 at 5:26
• 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. Apr 3 at 5:28
• 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) Apr 3 at 5:28
• 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 ? 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

• 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? Apr 4 at 1:56
• @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, Apr 4 at 20:35