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For the past two weeks I have been dealing with differential equations and I have stumbled upon a seemingly simple problem. Suppose I have a separable differential equation $\frac{dy}{dx}=y(x)$. This simply tells that the function value has to equal to the slope at a particular $x$ value. The solution is therefore simply $e^x$. So far so good. Now suppose I have similiar equation, namely $\frac{dy}{dx}=x^2$. Now this tells me that the slope of the function at a particular $x$ has to equal $x^2$. Hence, the solution is simply the integral, namely $\frac{1}{3}x^3$. One does not even need the conventional methods to solve separable differential equations.

However, let me now define $y(x)=x^2$. The two differential equations (and hence the solutions) should be now equivalent. As we have seen, that's not the case. I can sense that I am commiting a mathematical fallacy, however, I do not know what seems to be the problem when I solve the differential equation implicitly rather than explicitly as stated above. Could anyone help me please?

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  • $\begingroup$ I am not sure if I understand, but the point is that $y(x)=x^2$ does not fulfill any of those differential equations. Your first paragraph is indeed correct: the slope, not the function, is defined by the DE. $\endgroup$ – Miguel Dec 29 '15 at 11:44
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You found that a solution to $\frac{dy}{dx}=x^2$ is $y(x)=\frac{1}{3}x^3$, so now when you set $y(x)=x^2$, you change $y(x)$ from the previous $y(x)=\frac{1}{3}x^3$ and it is no longer a solution to the equation. The new $y(x)$ is a solution to $\frac{dy}{dx}=2x$, because $\frac{d}{dx}(x^2)=2x$. The equation $\frac{dy}{dx}=y(x)$ is only satisfied by a function that is equal to its derivative everywhere and neither $\frac{1}{3}x^3$ nor $x^2$ are such functions. Family of functions $y(x)=Ce^x$, $C \in \mathbb{R}$, are only solutions.

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