First Order Linear Equations Integrating Factors

You could solve for y using separation of variables to get But what about this one? Wait! We can’t separate them There is a way that starts with remembering the Product Rule for Derivatives…

f  g + g f g = f = I(x) f = I(x)P(x) g = y I(x) = I(x)P(x) A first order linear differential equation is of the form: …where P and Q are continuous functions. Notice that the equation is not separable. We are going to have to find another way to solve for y Let’s multiply the equation by a new—and continuous—function I(x): Why do this? f  g + g f It allows us to label the left side of the equation in product rule terms like this: g = f = I(x) f = I(x)P(x) g = y I(x) = I(x)P(x) Which also implies that

g f Now if we can just find a general formula for I(x), we can work the product rule backwards. We don’t need to worry about a constant of integration until after we solve the differential equation which is also why we don’t need a when we dump the absolute value signs. Therefore, our integrating factor is…

So we multiply I(x) by the differential equation in order to solve the differential equation. Now let’s try the first one in which we already know the answer so we can test this method…

Solve for y First, identify P(x) and Q(x) Now, identify I(x) So now the equation becomes: By “undo-ing” the product rule on the left side, we get

Use substitution to solve this integral. Now let’s try the one in which we couldn’t separate the variables…

Solve for y First, identify P(x) and Q(x) Now, identify I(x) So now the equation becomes: By “undo-ing” the product rule on the left side, we get

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