Solve the Exact Differential Equation

Question

\begin{equation} (\frac{1}{x}+\frac{1}{x^2}-\frac{y}{x^2+y^2})dx+(ye^y+\frac{x}{x^2+y^2})dy=0 \end{equation} The above differential equation is exact. According to the text book I'm using the formula to solve this differential equation is the following: \begin{equation} f(x,y)=\int_{x_o}^{x}M(x,y)dx+\int_{y_o}^{y}N(x_o,y)dy=c \end{equation} In my lecture notes the solution is the following \begin{equation} ln(x)-\frac{1}{x}-arctan(\frac{x}{y})+ye^y-ey=C \end{equation} When trying to solve assuming $x_o$ be 1 and $y_o$ be 0 I have a remaining arctan function. If I had to guess it would be due to my point $x_o$ and $y_o$, and that it doesn t create a simply connected region.

Answer 1

The correct formula to solve the differential equation is $$ f(x,y)=\int_{x_0}^xM(x',y)\,dx'+\int_{y_0}^yN(x_0,y')\,dy'=C. \tag{1} $$ Proof. Using Leibniz integral rule in $(1)$, we obtain $$ \frac{\partial f}{\partial x}=M(x,y) \tag{2} $$ and \begin{align} \frac{\partial f}{\partial y}&=\int_{x_0}^x\frac{\partial}{\partial y}M(x',y)\,dx'+N(x_0,y) \\ &=\int_{x_0}^x\frac{\partial}{\partial x'}N(x',y)\,dx'+N(x_0,y) \\ &=N(x,y)-N(x_0,y)+N(x_0,y)=N(x,y).\quad\square \phantom{\int} \tag{3} \end{align} Now, let's apply $(1)$ to the given ODE: \begin{align} \int_{x_0}^xM(x',y)\,dx'&=\int_{x_0}^x\left(\frac{1}{x'}+\frac{1}{(x')^2}-\frac{y}{(x')^2+y^2}\right)dx' \\ &=\ln|x|-\frac{1}{x}-\arctan\left(\frac{x}{y}\right) -\ln|x_0|+\frac{1}{x_0}+\arctan\left(\frac{x_0}{y}\right), \tag{4} \\ \int_{y_0}^yN(x_0,y')\,dy'&=\int_{y_0}^y\left(y'e^{y'}+\frac{x_0}{x_0^2+(y')^2}\right)dy' \\ &=(y-1)e^y+\arctan\left(\frac{y}{x_0}\right)-(y_0-1)e^{y_0}-\arctan\left(\frac{y_0}{x_0}\right). \tag{5} \end{align} Substituting $(4)$ and $(5)$ in $(1)$, we get \begin{align} f(x,y)&=\ln|x|-\frac{1}{x}-\arctan\left(\frac{x}{y}\right)+(y-1)e^y \\ &\phantom{=}+\arctan\left(\frac{x_0}{y}\right)+\arctan\left(\frac{y}{x_0}\right)+g(x_0,y_0)=C. \tag{6} \end{align} Since $g(x_0,y_0)$ is a constant, and $$ \arctan\left(\frac{x_0}{y}\right)+\arctan\left(\frac{y}{x_0}\right) =\frac{\pi}{2}, \tag{7} $$ we can rewrite the solution to the ODE as $$ \ln|x|-\frac{1}{x}-\arctan\left(\frac{x}{y}\right)+(y-1)e^y=C', \tag{8} $$ where $C'$ is a new constant.