# 12.1: First Order Differential Equations


Before moving on, we first define an $$n$$-th order ordinary differential equation. It is an equation for an unknown function $$y(x)$$ that expresses a relationship between the unknown function and its first $$n$$ derivatives. One could write this generally as $F\left(y^{(n)}(x), y^{(n-1)}(x), \ldots, y^{\prime}(x), y(x), x\right)=0 .\label{eq:1}$ Here $$y^{(n)}(x)$$ represents the $$n$$th derivative of $$y(x)$$.

An initial value problem consists of the differential equation plus the values of the first $$n-1$$ derivatives at a particular value of the independent variable, say $$x_{0}$$ : $y^{(n-1)}\left(x_{0}\right)=y_{n-1}, \quad y^{(n-2)}\left(x_{0}\right)=y_{n-2}, \quad \ldots, \quad y\left(x_{0}\right)=y_{0} .\label{eq:2}$

A linear $$n$$th order differential equation takes the form $\left.a_{n}(x) y^{(n)}(x)+a_{n-1}(x) y^{(n-1)}(x)+\ldots+a_{1}(x) y^{\prime}(x)+a_{0}(x) y(x)\right)=f(x) .\label{eq:3}$ If $$f(x) \equiv 0$$, then the equation is said to be homogeneous, otherwise it is called nonhomogeneous.

Typically, the first differential equations encountered are first order equations. A first order differential equation takes the form $F\left(y^{\prime}, y, x\right)=0 .\label{eq:4}$ There are two common first order differential equations for which one can formally obtain a solution. The first is the separable case and the second is a first order equation. We indicate that we can formally obtain solutions, as one can display the needed integration that leads to a solution. However, the resulting integrals are not always reducible to elementary functions nor does one obtain explicit solutions when the integrals are doable.

## Separable Equations

A first order equation is separable if it can be written the form $\frac{d y}{d x}=f(x) g(y) \text {. }\label{eq:5}$ Special cases result when either $$f(x)=1$$ or $$g(y)=1$$. In the first case the equation is said to be autonomous.

The general solution to equation $$\eqref{eq:5}$$ is obtained in terms of two integrals: $\int \frac{d y}{g(y)}=\int f(x) d x+C,\label{eq:6}$ where $$C$$ is an integration constant. This yields a 1-parameter family of solutions to the differential equation corresponding to different values of $$C$$. If one can solve $$\eqref{eq:6}$$ for $$y(x)$$, then one obtains an explicit solution. Otherwise, one has a family of implicit solutions. If an initial condition is given as well, then one might be able to find a member of the family that satisfies this condition, which is often called a particular solution.

##### Example $$\PageIndex{1}$$

$$y^{\prime}=2 x y, y(0)=2$$.

###### Solution

Applying $$\eqref{eq:6}$$, one has $\int \frac{d y}{y}=\int 2 x d x+C .\nonumber$ Integrating yields $\ln |y|=x^{2}+C .\nonumber$ Exponentiating, one obtains the general solution, $y(x)=\pm e^{x^{2}+C}=A e^{x^{2}} .\nonumber$ Here we have defined $$A=\pm e^{C}$$. Since $$C$$ is an arbitrary constant, $$A$$ is an arbitrary constant. Several solutions in this 1-parameter family are shown in Figure $$\PageIndex{1}$$.

Next, one seeks a particular solution satisfying the initial condition. For $$y(0)=$$ 2 , one finds that $$A=2$$. So, the particular solution satisfying the initial condition is $$y(x)=2 e^{x^{2}}$$.

##### Example $$\PageIndex{2}$$

$$y y^{\prime}=-x$$.

###### Solution

Following the same procedure as in the last example, one obtains: $\int y d y=-\int x d x+C \Rightarrow y^{2}=-x^{2}+A, \text { where } A=2 C .\nonumber$ Thus, we obtain an implicit solution. Writing the solution as $$x^{2}+y^{2}=A$$, we see that this is a family of circles for $$A>0$$ and the origin for $$A=0$$. Plots of some solutions in this family are shown in Figure $$\PageIndex{2}$$.

## Linear First Order Equations

The second type of first order equation encountered is the linear first order differential equation in the standard form $y^{\prime}(x)+p(x) y(x)=q(x) .\label{eq:7}$ In this case one seeks an integrating factor, $$\mu(x)$$, which is a function that one can multiply through the equation making the left side a perfect derivative. Thus, obtaining, $\frac{d}{d x}[\mu(x) y(x)]=\mu(x) q(x) .\label{eq:8}$

The integrating factor that works is $$\mu(x)=\exp \left(\int^{x} p(\xi) d \xi\right)$$. One can derive $$\mu(x)$$ by expanding the derivative in Equation $$\eqref{eq:8}$$, $\mu(x) y^{\prime}(x)+\mu^{\prime}(x) y(x)=\mu(x) q(x),\label{eq:9}$ and comparing this equation to the one obtained from multiplying $$\eqref{eq:7}$$ by $$\mu(x)$$ : $\mu(x) y^{\prime}(x)+\mu(x) p(x) y(x)=\mu(x) q(x) .\label{eq:10}$ Note that these last two equations would be the same if the second terms were the same. Thus, we will require that $\frac{d \mu(x)}{d x}=\mu(x) p(x) .\nonumber$ This is a separable first order equation for $$\mu(x)$$ whose solution is the integrating factor: $\mu(x)=\exp \left(\int^{x} p(\xi) d \xi\right) . \label{eq:11}$

Equation $$\eqref{eq:8}$$ is now easily integrated to obtain the general solution to the linear first order differential equation: $y(x)=\frac{1}{\mu(x)}\left[\int^{x} \mu(\xi) q(\xi) d \xi+C\right] .\label{eq:12}$

##### Example $$\PageIndex{3}$$

$$x y^{\prime}+y=x, \quad x>0, y(1)=0$$.

###### Solution

One first notes that this is a linear first order differential equation. Solving for $$y^{\prime}$$, one can see that the equation is not separable. Furthermore, it is not in the standard form $$\eqref{eq:7}$$. So, we first rewrite the equation as $\frac{d y}{d x}+\frac{1}{x} y=1 \text {. }\label{eq:13}$ Noting that $$p(x)=\frac{1}{x}$$, we determine the integrating factor $\mu(x)=\exp \left[\int^{x} \frac{d \xi}{\xi}\right]=e^{\ln x}=x .\nonumber$ Multiplying equation $$\eqref{eq:13}$$ by $$\mu(x)=x$$, we actually get back the original equation! In this case we have found that $$x y^{\prime}+y$$ must have been the derivative of something to start. In fact, $$(x y)^{\prime}=x y^{\prime}+x$$. Therefore, the differential equation becomes $(x y)^{\prime}=x .\nonumber$ Integrating, one obtains $x y=\frac{1}{2} x^{2}+C\nonumber$ or $y(x)=\frac{1}{2} x+\frac{C}{x} .\nonumber$

Inserting the initial condition into this solution, we have $$0=\frac{1}{2}+C$$. Therefore, $$C=-\frac{1}{2}$$. Thus, the solution of the initial value problem is $y(x)=\frac{1}{2}\left(x-\frac{1}{x}\right) .\nonumber$

We can verify that this is the solution. Since $$y^{\prime}=\frac{1}{2}+\frac{1}{2 x^{2}}$$, we have $x y^{\prime}+y=\frac{1}{2} x+\frac{1}{2 x}+\frac{1}{2}\left(x-\frac{1}{x}\right)=x \text {. }\nonumber$ Also, $$y(1)=\frac{1}{2}(1-1)=0$$.

##### Example $$\PageIndex{4}$$

$$(\sin x) y^{\prime}+(\cos x) y=x^{2}$$.

###### Solution

Actually, this problem is easy if you realize that the left hand side is a perfect derivative. Namely, $\frac{d}{d x}((\sin x) y)=(\sin x) y^{\prime}+(\cos x) y .\nonumber$ But, we will go through the process of finding the integrating factor for practice.

First, we rewrite the original differential equation in standard form. We divide the equation by $$\sin x$$ to obtain $y^{\prime}+(\cot x) y=x^{2} \csc x\nonumber$ Then, we compute the integrating factor as $\mu(x)=\exp \left(\int^{x} \cot \xi d \xi\right)=e^{\ln (\sin x)}=\sin x .\nonumber$ Using the integrating factor, the standard form equation becomes $\frac{d}{d x}((\sin x) y)=x^{2} .\nonumber$ Integrating, we have $y \sin x=\frac{1}{3} x^{3}+C\nonumber$ So, the solution is $y(x)=\left(\frac{1}{3} x^{3}+C\right) \csc x .\nonumber$

This page titled 12.1: First Order Differential Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.