First Order Differential Equations - Problem Solving

First order differential equations are useful because of their applications in physics, engineering, etc. They can be linear, of separable, homogenous with change of variables, or exact. Each one has a structure and a method to be solved.

A homogenous equation with change of variables needs to be in the form $$\frac{dy}{dx}=F(x,y),$$ with a function $F(x,y)$ such that $F(tx,ty)=F(x,y) \Rightarrow F(x,y)=F\big(1,\frac{y}{x}\big)$. So we can do the change of variables $u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u$, getting the equation (most of the time, separable) $$\frac{du}{dx}x+u=F(1,u).$$

Solve the differential equation $$\frac{dy}{dx}=\frac{y(y+x)}{x^2}.$$

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At first we can find out that this equation is said to be homogenous: $$\frac{dy}{dx}=\frac{y(y+x)}{x^2}=\frac{ty(ty+tx)}{t^2x^2}=\frac{t^2y(y+x)}{t^2x^2}=\frac{y(y+x)}{x^2}.$$ Then, we can do the change of variables: $$u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u,$$ and get the following separable equation: $$\begin{aligned} \frac{du}{dx}x+u&=u^2+u \\\\ \frac{du}{dx}&=\frac{u^2}{x} \\\\ \int \frac{du}{u^2}&=\displaystyle \int \frac{dx}{x}\\\\ \frac{1}{u}&=-\ln(x)+C \\\\ u&=-\frac{1}{\ln(x)+C} \\\\ y&=-\frac{x}{\ln(x)+C}, \end{aligned}$$ where $C$ is the constant of integration. $_\square$

Let's take the following differential equation: $$\begin{aligned} A(x,y)dy+B(x,y)dx&=0\\ \Rightarrow \frac{dy}{dx}&=-\frac{B(x,y)}{A(x,y)}. \end{aligned}$$ It would be a very nice coincidence that there was a function $F(x,y)$ such that $$\frac{dy}{dx}=-\frac{\frac{\partial F(x,y)}{\partial x}}{\frac{\partial F(x,y)}{\partial y}}=-\frac{\partial y}{\partial x}.$$ A differential equation is said to be exact iif there exists a $C^2$ function $F(x,y)$ such that $A(x,y)=\frac{\partial F(x,y)}{\partial y}$ and $B(x,y)=\frac{\partial F(x,y)}{\partial x}$. $$$$ To prove that such a function exists we can use partial derivatives: $$\begin{aligned} A(x,y)=\frac{\partial F(x,y)}{\partial y} &\Rightarrow \frac{\partial A(x,y)}{\partial x}=\frac{\partial ^2 F(x,y)}{\partial x \partial y}\\ B(x,y)=\frac{\partial F(x,y)}{\partial x} &\Rightarrow \frac{\partial B(x,y)}{\partial y}=\frac{\partial ^2 F(x,y)}{\partial y \partial x}. \end{aligned}$$ As $F$ is $C^2,$ we can conclude $$\begin{aligned} \frac{\partial ^2 F(x,y)}{\partial y \partial x}&=\frac{\partial ^2 F(x,y)}{\partial x \partial y}\\ \Rightarrow \frac{\partial A(x,y)}{\partial x}&=\frac{\partial B(x,y)}{\partial y}. \end{aligned}$$ And that is the most important requirement for saying a differential equation is exact. $$$$ When the function $F$ is found, the solution to the differential equation is given, implicitly, by $$F(x,y)=0.$$

Solve the differential equation $$\ln\big(x^2y\big)dy+\frac{2\ln(y)}{x}dx=0.$$

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On the way to prove that is an exact equation is to take $$\begin{aligned} \frac{\partial}{\partial x}\big(\ln(x^2y)\big)&=\frac{\partial}{\partial y}\left(\frac{2\ln(y)}{x}\right)\\ \frac{2}{xy}&=\frac{2}{xy}. \end{aligned}$$ So it is exact, and we know $$\begin{aligned} \ln(x^2y)=\frac{\partial F(x,y)}{\partial y} \Rightarrow F(x,y) &= \int \ln(x^2y)dy \\ &=y\ln(x^2y)-\displaystyle \int dy\\ &=y\big(\ln(x^2y)+1\big)+C(x) \end{aligned}$$ and $$\begin{aligned} \frac{2\ln(y)}{x}&=\frac{\partial F(x,y)}{\partial x} \\ \frac{2\ln(y)}{x}&=\frac{2}{x}+C'(x)\\ C(x)&=2\ln(x)\ln(y)-2\ln(x)+C\\ F(x,y)&=y\big(\ln(x^2y)+1\big)+2\ln(x)\big(\ln(y)+1\big)+C. \end{aligned}$$

So the solution to the differential equation is given by

$$y\big(\ln(x^2y)+1\big)+2\ln(x)\big(\ln(y)+1\big)+C=0.\ _\square$$