Integrating Factors

Integrating factors allow us to solve equations of the form $y^{\prime }+P(x)y=Q(x)$

Intuition

The derivative of $e^{R(x)}$ is $R^{\prime }(x)e^{R(x)}$ . This can be used to cancel out terms when we have anything of the form $y^{\prime }+R^{\prime }(x)y$, and we can do so by setting $P(x)=R^{\prime }(x)$ . By playing around with this idea the methodology below was developed.

General Solutions initially defined

We know that $y^{\prime }+P(x)y=Q(x)$. We want $R^{\prime }(x)=P(x)$, so we define $R(x)=\int P(x)dx$ to get $y^{\prime }+R^{\prime }(x)y=Q(x)$. By multiplying all terms of the equation by $e^{R(x)}$ we get

$$\begin{array}{rl}Q(x)e^{R(x)}& =y^{\prime }{e}^{R(x)}+y{R}^{\prime }(x)e^{R(x)}\\ & =y^{\prime }{e}^{R(x)}+y(e^{R(x)}{)}^{\prime }\\ & =(y{e}^{R(x)}{)}^{\prime }\end{array}$$

By integrating both sides and then rearranging for $y$, we get $y{e}^{R(x)}=\int Q(x)e^{R(x)}dx\iff y=e^{-R(x)}\int Q(x)e^{R(x)}dx$with $R(x):=\int P(x)dx$.

Example

Many examples require harder integrals than the one that follows, however this example should give you an idea of how to proceed. $y^{\prime }-\frac{y}{x}=x$ We have that $P(x)=-\frac{1}{x}$, so $R(x)=\int P(x)dx=-\int \frac{1}{x}dx=-\ln x$. Hence our Integrating Factor is $e^{-\ln x}=\frac{1}{x}$, so we can rewrite our original equation by multiplying it by $\frac{1}{x}$ as

$$\begin{array}{rl}\frac{y^{\prime }}{x}-\frac{y}{x}\frac{1}{x}& =x\frac{1}{x}\\ \frac{y^{\prime }}{x}-\frac{y}{x^2}& =1\\ \frac{y^{\prime }}{x}-y{\left(\frac{1}{x}\right)}^{\prime }& =1\\ {\left(\frac{y}{x}\right)}^{\prime }& =1\end{array}$$

Integrating both sides with respect to $x$ we get $\frac{y}{x}=\int 1dx=x\iff y=x^2$ . Confirming it within the original equation, we indeed get $y^{\prime }-\frac{y}{x}=(x^2{)}^{\prime }-\frac{x^2}{x}=2x-x=x$.