Extensão: Integração entre fatores

A first-order linear differential equation is an equation that can be put into the following form: $$\frac{dy}{dx} + P(x)y = Q(x)$$

As an example, consider the following differential equation, taken from James Stewart's Calculus: Early Transcendentals, 6th Edition: $$y' + \frac{1}{x}y = 2$$ This equation is not separable, as it is impossible to put $dy$ and expressions with $y$ entirely on one side of the equation, while putting $dx$ and expressions with $x$ entirely on the other side. However, a multiplication by $x$ yields: $$xy' + y = 2x$$ which (if you are very perceptive) you might recognize as: $$(xy)' = 2x$$ by the Product Rule for Derivatives. Therefore, $xy = x^2 + C$, so $y = x + \frac{C}{x}$.

It turns out that all first-order linear differential equations can be solved in this way (not all of them need to be done this way, however). The factor by which we multiply both sides is called an integrating factor.

Consider a differential equation of the form: $$\frac{dy}{dx} + P(x)y = Q(x)$$ We wish to find an integrating factor $I(x)$ such that: $$I(x)(y' + P(x)y) = (I(x)y)'$$ Expanding and using the Product Rule, $$I(x)y' + I(x)P(x)y = I'(x)y + I(x)y'$$ $$I(x)P(x) = I'(x)$$

We see that this is a separable differential equation in $I(x)$, so it can be easily solved. You are encouraged to try to solve this differential equation yourself. What did you find for $I(x)$? Continue reading, and see if you are correct!

We now solve the differential equation: $$I(x)P(x) = \frac{dI}{dx}$$ We want $I$ to be nonzero, since it should be a nontrivial integrating factor, so the next step is permitted: $$\frac{dI}{I} = P(x)dx$$ $$\int\frac{dI}{I} = \int P(x)dx$$ $$\ln|I| + C = \int P(x)dx$$ $$I(x) = \pm e^{-C} e^{\int P(x)dx}$$ Letting $A = \pm e^{-C}$ (this is an arbitrary nonzero constant), we have: $$I(x) = Ae^{\int P(x)dx}$$ We only need a particular integrating factor, so we let $A=1$ for simplicity: $$I(x) = e^{\int P(x)dx}$$ This is our integrating factor.

Now, suppose that we have such an integrating factor $I(x)$. Let us see how we will solve the differential equation.

We have the equation: $$\frac{dy}{dx} + P(x)y = Q(x)$$ Multiplying both sides by $I(x)$, we have: $$I(x)(y' + P(x)y) = I(x)Q(x)$$ $$(I(x)y)' = I(x)Q(x)$$ Integrating both sides, $$I(x)y = \int I(x)Q(x)dx + C$$ $$y = \frac{1}{I(x)}\left(\int I(x)Q(x)dx + C\right)$$ Assuming that the function $I(x)Q(x)$ is integrable, we can find $y$, so we have solved the differential equation.