1
$\begingroup$

In separable differential equations of the form:

$ M(x,y) dx + N(x,y) dy =0$

We can separate variables and then integrate both sides to get the solution $y = f(x)$ to the differential equation.

In such an equation, when integrating, we cannot consider $y$ constant because what we're looking for is a solution curve in which $y$ is dependent on $x$ which has slope $ \frac {dy}{dx}$.

On the other hand, if we have an exact ( lets just assume all the conditions are met for it to be exact) equation of the same form,

$ M(x,y) dx + N(x,y) dy =0$

What we do is try to find a function $ \psi(x,y) = c$ that satisfies the exact differential equation, but the difference here ( and that's where I'm confused ) is that when we integrate for example,

$ M(x,y) = \left(\frac{\partial \psi}{\partial x}\right)$

When we integrate that last expression, we hold $y$ constant! Why are we doing that? Is it because we're solving for an implicit function in $x$ and $y$ [ $ \psi = c]$? But if we're solving for an implicit function then what's the meaning of $ \frac{dy}{dx}$ then? I hope my question was clear, thanks.

$\endgroup$
1
$\begingroup$

We hold $y$ constant because we are solving for differential equation $M(x,y)=\psi_x(x,y) = \frac{\partial\psi}{\partial x}$ where $\psi_x$ is the partial derivative of $\psi$ in respect to $x$. If integrating in respect one variable (in this case, $x$), we treat all other variables as constants, a lesson learned in multivariable calculus.

I also want to stress that the form $M(x,y)dx + N(x,y)dy = 0$ does not give the best picture of what is going on. Another way to think about exact equations is in the form $$\psi_x(x,y) + \psi_y(x,y)\frac{dy}{dx} = 0$$ We can work backwards from the form $\psi(x,y) = 0$ to get to here. We begin by treating $y$ as some function of $x$, which we will call $y(x)$. Then using the chain rule from multivariable calculus we can derive the following: \begin{align*} \frac{d}{dx}\psi(x,y(x)) &= \frac{d}{dx}C \\ \psi_x(x,y)+\frac{d\psi}{dy}\frac{dy}{dx} &= 0 \\ \psi_x(x,y) + \psi_y(x,y)\frac{dy}{dx} &= 0 \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.