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I calculated the partial differential of $y^2$ with respect to $x$. I thought this would be $0$ since no $x$ is present in the equation, so $x$ must be $1$. The derivative of $1$ is $0$, so $y^2*0 = 0$. The answer is $y^2$ though.

Why is this partial derivative not $0$? What am I missing?

The complete equation is $\dfrac{dm}{dx} = y^2$

context: This is a calculation to find if the differential equation $xydx+y^2dy=0$ is exact or not. I am calculating $\dfrac{dm}{dy}$ and $\dfrac{dm}{dx}$ to see if it is exact. The calculation above is $\dfrac{dm}{dx}$ .

user112167
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  • You might need to provide more context. For example, what other variables does $y$ depend on? Do any of these variables also depend on $x$? – Doug Aug 10 '22 at 19:01
  • Still not enough context. We know that the $x$-derivative of $m$ is $y^2$. No clue about the $x$-derivative of $y^2$ which would obviously be the second $x$-derivative of $m$. – Kurt G. Aug 10 '22 at 19:17
  • To check whether the ODE is exact or not, you should check $\dfrac{\partial {M}}{\partial{y}}=\dfrac{\partial {N}}{\partial{x}}$. Not $\dfrac{dM}{dy}$ – Manjoy Das Aug 10 '22 at 19:20
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    In this case, $M(x,y)=xy, N(x,y)=y^2$. You need $M_y=N_x$ for pde to be exact. $N_x=0$ – Vasili Aug 10 '22 at 19:25
  • @Vasili yes, but Nx is $y^2$ in the book. Is this a misprint? – user112167 Aug 10 '22 at 19:28
  • @user112167 . $N(x,y)$ is $y^2$ (no misprint). $N_x$ is an abbreviation for $\frac{\partial N}{\partial x}$ and that is ? Can you calculate this ? – Kurt G. Aug 10 '22 at 19:31
  • @KurtG. Yes, I calculated it to be 0 since the derivative of a constant is 0. It seemed to me the book was saying this derivative was $y^2$ itself and that my answer was wrong. – user112167 Aug 10 '22 at 19:47
  • If you want to get to the bottom of what this book was saying please edit your question one more time and give as many details as possible. – Kurt G. Aug 10 '22 at 19:51

1 Answers1

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Here $M=xy$ and $N=y^2$.
$\dfrac{\partial{M}}{\partial{y}}=x$ and $\dfrac{\partial{N}}{\partial{x}}=0$.
So ODE is not exact.

Note:
1) Obviously the partial derivative of $y^2$ w.r.t $x$ is $0$.

Manjoy Das
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