4.5 First-order linear equations  (Page 3/10)

Here $C_2$ can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of [link] by the integrating factor     $\mu \left(x\right).$ This gives

The left-hand side of [link] can be rewritten as $\frac{d}{dx}\left(\mu \left(x\right)y\right).$

Next integrate both sides of [link] with respect to $x.$

Divide both sides of [link] by $\mu \left(x\right)\text{:}$

Since $\mu \left(x\right)$ was previously calculated, we are now finished. An important note about the integrating constant $C\text{:}$ It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving $p\left(x\right)$ is necessary in order to find an integrating factor for [link] . Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for $C$ for this integral. We chose $C=0.$ When calculating the integral inside the brackets in [link] , it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to [link] . This integrating factor guarantees just that.

Now we use the same strategy to find the solution to an initial-value problem.

Applications of first-order linear differential equations

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.