Extraordinary Differential Equations #4

Relevant wiki: First Order Differential Equations - Problem Solving

Because all terms in the ODE are a combination of functions on both $x$ and $y(x)$ let's proceed to check if the ODE is in exact form. To do so, convert the ODE into the form $p(x,y)+q(x,y)y'=0$ where $p(x,y)=\frac{\partial F}{\partial x}$ and $q(x,y)=\frac{\partial F}{\partial y}$ for some function $F=F(x,y)$ . We test if a such function $F$ exists by checking if $\frac{\partial p}{\partial y}=\frac{\partial p}{\partial x}$ .

With some work we can determine that in fact $\frac{\partial}{\partial y} (-\frac{2017\ln{y}}{x^2}-e^{\sin{x}}\cos{x})=\frac{\partial}{\partial x} \frac{2017}{xy}=-\frac{2017}{x^2 y}$ . Then we get $\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} \frac{dy}{dx}$ =0 which implies that the general solution to this is in the form $F(x,y)=c$ where $c$ is the arbitrary constant of integration. To obtain $F(x,y)$ we integrate both $p(x,y)$ and $q(x,y)$ in terms of $x$ and $y$ respectively:

$F(x,y)=\int p(x,y) dx + k(y)=\frac{2017\ln{y}}{x}-e^{\sin{x}}+k(y)$

$F(x,y)=\int q(x,y) dy + h(x)=\frac{2017\ln{y}}{x}+h(x)$

By comparison of terms, $h(x)=-e^{\sin{x}}$ and $k(y)=0$ . We thus get:

$F(x,y)=\frac{2017\ln{y}}{x}-e^{\sin{x}}=c'$ $\ln{y}=\frac{c'x}{2017}+\frac{xe^{\sin{x}}}{2017}$ $y=e^{cx+\frac{xe^{\sin{x}}}{2017}}$