# Thread: Exactness Test of differential equations

1. ## Exactness Test of differential equations

I have the problem of (ylny + e^x)dx + (x+e^-y + xlny)dy. I have determined that they are exact. However, I am running into a problem when I am trying to solve the equation. I have done the following and got stuck!

1. partial F with respect to x = ylny + e^x and
2. partial F with respect to y = x + e^-y + xlny

When I integrated #1 with respect to x, I came up with: xylny + e^x + g(y) and then when I integrated #2 with respect to y, I came up with:
xy - e^-y + xylny -x + h(x). From here I am stuck, therefore I think I made a mistake somewhere and I can't seem to find it. Can somebody help?

2. $(y\ln{y} + e^x)dx + (x+e^{-y} + x\ln{y})dy=0$

Set $M(x,y)=y\ln{y} + e^x$ and $N(x,y)=x+e^{-y} + x\ln{y}$

You found an exact ODE, therefore, exists a function $f(x,y)$ such that

$\frac{\partial{f}}{\partial{x}}=y\ln{y} + e^x$ and $\frac{\partial{f}}{\partial{y}}=x+e^{-y} + x\ln{y}$

Integrate the first equation with respect to $x$, after that find the partial derivative with respect to $y$.

3. Hello, graceofayak!

I have the problem of: . $(y\ln y + e^x)dx + (x + e^{-y} + x\ln y)dy \:=\:0$

I have determined that they are exact.
I have done the following and got stuck:

$(1)\;\frac{\partial f}{\partial x} \:=\:y\ln y + e^x$ .and . $(2)\;\frac{\partial f}{\partial y} \:=\: x + e^{-y} + x\ln y$

When I integrated #1 with respect to x, I came up with: $xy\ln y + e^x + g(y)$ . Yes!

Then when I integrated #2 with respect to y, I came up with: . $xy - e^{-y} + xy\ln y -x + h(x)$ . no

Your error is in integrating the third term: . $x\int \ln y\,dy$

By parts: . $\begin{Bmatrix}u & = & \ln y & \quad & dv & = & dy \\ du & = & \frac{dy}{y} & \quad & v & = & y\end{Bmatrix}$

Then: . $x\left(y\ln y - \int dy\right) \;=\;x\left(y\ln y - y\right)\;=\;xy\ln y - xy$

So the integral of #2 is: . $xy - e^{-y} + xy\ln y - xy + h(x)$