# Seperation of Variables/Integrating Factor Method

## Homework Statement

The first order linear equation of the form:
$\frac{dy}{dx} + ay = b$
where a and b are constants, can be solved both by the integrating factor method and by seperation of variables. Solve the equation using both methods to see that you get the same solution.

## Homework Equations

Separation of variables and Integrating factor equations

## The Attempt at a Solution

I am 80% sure that I have correctly found the solution using both methods; however, I am having trouble equating the two solutions. This is what I have done so far, first I will show using the integrating factor method, followed by separation of variables:
$p(x) = a$
$u(x) = e^{ax}$
$\frac{d}{dx}(e^{ax} y(x)) = be^{ax}$
$\int\frac{d}{dx}(e^{ax} y(x))\,dx = \int be^{ax}\, dx$
$e^{ax} y(x) = abe^{ax} + C$
$y(x) = ab + \frac{C}{e^ax}$
Now again using the separation of variables method:
$\frac{dy}{dx} = b-ay$
$\frac{dy}{y} = (b-a)dx$
$\int\frac{dy}{y} = \int b-a\, dx$
$ln(y) = bx - ax + C$
$y = e^{bx} - e^{ax} + e^C$

(b-ay)/y =b/y -a

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

The first order linear equation of the form:
$\frac{dy}{dx} + ay = b$
where a and b are constants, can be solved both by the integrating factor method and by seperation of variables. Solve the equation using both methods to see that you get the same solution.

## Homework Equations

Separation of variables and Integrating factor equations

## The Attempt at a Solution

I am 80% sure that I have correctly found the solution using both methods; however, I am having trouble equating the two solutions. This is what I have done so far, first I will show using the integrating factor method, followed by separation of variables:
$p(x) = a$
$u(x) = e^{ax}$
$\frac{d}{dx}(e^{ax} y(x)) = be^{ax}$
$\int\frac{d}{dx}(e^{ax} y(x))\,dx = \int be^{ax}\, dx$
$e^{ax} y(x) = abe^{ax} + C$
$y(x) = ab + \frac{C}{e^ax}$
Now again using the separation of variables method:
$\frac{dy}{dx} = b-ay$
$\frac{dy}{y} = (b-a)dx$
$\int\frac{dy}{y} = \int b-a\, dx$
$ln(y) = bx - ax + C$
$y = e^{bx} - e^{ax} + e^C$

Never write ##1/e^{ax}##; always write it as ##e^{-ax}##, because that way you can apply standard differentiation/integration formulas.

Your second method has an obvious blunder: from
$$\frac{dy}{dx} = b-ay$$
$$\frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}$$

Never write ##1/e^{ax}##; always write it as ##e^{-ax}##, because that way you can apply standard differentiation/integration formulas.

Your second method has an obvious blunder: from
$$\frac{dy}{dx} = b-ay$$
$$\frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}$$