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Given the following example:

$$\frac{dy}{dx} = \frac{x^2}y$$

I understand how one would go about solving this. However, I do not understand one thing.

If the original equation was the following: $$\frac{dy}{dx} = \frac{x^2}{y}$$, then this means that y is the original function that we are trying to find.

Then going on to solve this, we obtain: $$\int x^2dx = \int y\,dy$$

Does this then mean that we are integrating $y$, a function, with respect to itself? If so, what does integrating a function with respect to itself mean? I'm used to integrating a function with respect to a variable, and am somewhat stuck on the meaning of this. I really appreciate any help on this. Thanks so much.

Another User
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Daanyal Ali Akhtar
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  • Let $y$ be a function of $x$, i.e. $y(x)$. The chain rule states that $[f(y(x))]' = f'(y(x))\cdot y'(x)$, and $\int f'(y(x)) dy(x) = f(y(x))+C$. Here you have $x^2=f'(y(x))\cdot {dy(x)\over dx}$ or equivalently $x^2 dx = f'(y(x)) dy(x)$, where $f'(y(x))=y(x)$. – David P Aug 20 '22 at 19:03

2 Answers2

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$y$ has to be read as $y(x)$. In your case it's like:

$$y'(x) = \frac{x^2}{y(x)}$$

Going through the differentials, it would be solvabe by separation of variables:

$$\frac{\text{d}y}{\text{d}x} = \frac{x^2}{y} \longrightarrow y\ \text{d}y = x^2\ \text{d}x$$

Now you integrate indeed:

$$\frac{y^2}{2} = \frac{x^3}{3} +C$$

The constant may be set to zero, according to some initial condition (we just do it here). Then:

$$y(x) = \sqrt{\frac{2}{3}x^3}$$

Notice that it's indeed what you want, for

$$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x} \sqrt{\frac{2}{3}x^3} = \sqrt{\frac{3}{2x^3}}x^2 \equiv \frac{x^2}{\sqrt{\frac{2}{3}x^3}} \equiv \frac{x^2}{y(x)}$$

Enrico M.
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$$\int y d(y) = \frac{y^{2}}{2} + C$$ by the rule for integrating polynomials. Of course, $y$ is a function but it can be treated as a variable. Moreover, notice the following.

$$\int y d(y) = \int y y'dx = y^{2} - \int yy'dx. $$

This also works as the differential $dy$ is given by $y'dx$ since $y$ depends on $x$.

Look up the Riemann-Stieltjes integral.

Artur Wiadrowski
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