Derivation of Expressions
The following section shows how we arrive at the expressions of solving \(\mu\) and the first-order linear DE general solution. We derive both by making parallelism with two equations: (1) the modified standard form and (2) the product rule.
To begin, let's create the modified standard form. Multiply the standard form, \(y^\prime + a(x)y = b(x)\), with \(\mu\):
- \(\mu \cdot \frac{d y}{d x}+y \mu a(x)=\mu b(x)\)
Next, we consider the product rule, which states that:
- \(\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\)
Let's make some parallelism with these two expressions:
- \(\mu = u\)
- \(\frac{dy}{dx} = \frac{dv}{dx}\)
- \(y = v\)
- \(\mu a(x) = \frac{du}{dx}\)
Now that we've established the parallels between the two expressions, we can derive the integrating factor and the general solution.
Integrating Factor
Consider the fourth expression: \(\mu a(x) = \frac{du}{dx}\). We can rewrite this as \(\mu a(x) = \frac{d\mu}{dx}\) because \(\mu = u\) (first expression). It is now a differential equation that can be solved using separation of variables.
If we put \(\mu\) on the left side of the equation and all \(x\) on the other side, we can integrate both sides to form the expression of the integrating factor.
- \(\mu a(x) = \frac{d\mu}{dx}\)
- \(\frac{d \mu}{\mu}=a(x) d x\)
- \(\frac{d \mu}{\mu}=a(x) d x\)
- \(e^{\ln (\mu)=} e^{\int a(x) d x}\)
- \(\mu=e^{\int a(x) d x}\)
General Solution
Going back to our standard form and the parallelism, we can say that the expression, \(\mu \cdot \frac{d y}{d x}+y \mu a(x)=\mu b(x)\), is equal to \(\frac{d}{d x}(y \mu)\). From this, we can change the expression to:
- \(\frac{d}{d x}(y \mu)=\mu b(x)\)
We can then integrate both sides and isolate \(y\) to get the general solution expression.
- \(\int d(y \mu)=\int \mu b(x) d x\)
- \(y \mu=\int \mu b(x) d x\)
- \(y=\frac{1}{\mu} \int \mu b(x) d x\)
Example
Consider the differential equation: \(y^\prime + xy = x\) and solve for a general solution.
Investigate DE
First, we would investigate if this DE observes the form: \(y^\prime + a(x)y = b(x)\). In this case:
- \(a(x) = x\)
- \(b(x) = x\).
The form follows the standard format; hence, we can use the integrating factor and the general solution equation.
Apply Method
First, we find the integrating factor \(\mu\):
- \(\mu=e^{\int a(x) d x}\)
- \(\mu=e^{\int x d x}\)
- \(\mu=e^{\frac{x^2}{2}}\)
Finally, using the general solution expression, we obtain the following:
- \(y=\frac{1}{\mu} \int \mu b(x) d x\)
- \(y=\frac{1}{e^{\frac{x^2}{2}}} \int x e^{\frac{x^2}{2}} d x\)
The problem is now an integration problem. We solve for it to generate the general solution to the DE.
- \(y=\frac{1}{e^{\frac{x^2}{2}}}\left(e^{\frac{x^2}{2}}+C\right)\)
- \(y=1+C e^{\frac{-x^2}{2}}\)
Summary
An equation of the form \(y^\prime + a(x)y = b(x)\) is a first-order linear differential equation (DE) (also known as the standard form of a differential equation). The expressions \(\frac{dy}{dx}\), \(y^\prime\) is the first derivative, \(a(x)\) and \(b(x)\) are functions, and \(y\) is the dependent variable
We must apply two general expressions to solve this DE: integrating factor \(\mu\) and the general solution equation.
Integrating factor is equal to: \(\mu=e^{\int a(x) d x}\)
General solution equation is equal to: \(y=\frac{1}{\mu} \int \mu b(x) d x\)
We obtained these equations by making a parallelism with the modified standard form (standard form multiplied by integrating factor) and the product rule.
From this parallelism, the solution to a first-order linear differential equation is \(\frac{d}{d x}(y \mu)=\mu b(x)\)
We derive the integrating factor \(\mu\) from the parallels between the two expressions and use the separation of variables.
