# Help solving a D. E using Linear Equation solving method

I have the following equation and need to use linear equation method to solve this

$$y' + 5y = 2$$

(Do not use separable equation method for this please)

• Is it clear to you what the "method" is? The answer could be instructor dependent, but probably is not. You need to (a) find the general solution of the homogeneous equation $y'+5y=0$, probably by trying $y=e^{rx}$ and (b) find a particular solution of $y'+5y=2$ (easy). – André Nicolas Jul 11 '11 at 4:01
• "Linear equation method" means to use a certain formula? What is that formula? – GEdgar Jul 11 '11 at 4:02
• Integrating factors? I don't know what you mean by this question at all... – mathmath8128 Jul 11 '11 at 5:01
• I think we just have to wait until user10695 comes back and tells us what he means by "linear equation method". – GEdgar Jul 11 '11 at 13:15

## 1 Answer

The question suprises me a little, since it seems more basic than other DE questions that you have asked in the past. I will give an answer so as not to leave it unanswered. There have been a number of posts on linear differential equations on this site, but they all seem to involve equations that are harder than this one. Perhaps it is worthwhile to show how things work in a very simple case.

The given equation is a first order linear differential equation with constant coefficients. A general procedure for solving linear differential equations goes as follows.

($1$.) Find the general solution $y_g$ of the differential equation obtained by setting the term not containing $y$, $y'$ (and $y''$, $y'''$, and so on if they occur) equal to $0$. So in our case we want to find the general solution of the DE $y'+5y=0$.

($2$.) Find a particular solution $y_p$ of the original equation.

Then the general solution of our equation is $y_p +y_g$.

A particular solution: In this case, finding a particular solution is easy. We look for a constant function $y_p$, because then we will have $y_p'=0$. So we want $5y_p=2$, giving $y_p=\frac{2}{5}$.

The general solution of the homogeneous equation: The usual procedure is to look for solutions of the shape $y=e^{rx}$, where $r$ is a constant. In our case, we want to have $$y'+5y=0.$$ Let $y=e^{rx}$ and calculate. We obtain $$re^{5x}+5e^{rx}=0.$$ This equation holds identically precisely if $r+5=0$. So we conclude that $r=-5$.

Thus $y=e^{rx}$ is a solution of the homogeneous DE. The general solution of the homogeneous DE is given by $$y_g=Ce^{-5x},$$ where $C$ is any constant.

Thus the general solution of the original DE is given by $$y=\frac{2}{5}+Ce^{-5x}.$$

Remark: The same basic calculation, "particular solution plus general solution of the homogeneous DE" is valid for general linear differential equations. The details can become considerably harder.