Seperation of Variables/Integrating Factor Method

[_N3WTON_]

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 separation 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

 

[dirk_mec1]

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

 

[Ray Vickson]

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 separation 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
it dos NOT follow that
\frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}

 

[_N3WTON_]

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
it dos NOT follow that
\frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}

im a little confused about what is wrong with the separation I performed?
Edit: Nvm I see it now, thank you