Separation of Variables for ODE

[spaghetti3451]

Homework Statement

Solve the following equation by separation of the variables:

y' tan^-1x - y (1+x²)^-1 = 0

Homework Equations

The Attempt at a Solution

I am not sure if tan^-1x stands for arctan x or (tan x)^-1. (This has been taken out a book.) Any help on this would be appreciated.

If arctan x, then we have to integrate 1 / [(1+x²)(arctan x)] w.r.t x. No idea how to do this.

If (tan x)^-1, then we have to integrate tan x / (1+x²) w.r.t. x. No idea on how to do this either.

 

[SadScholar]

It's should be arctan(x). So
$$
y'\arctan(x)-\frac{y}{1+x^2}=0\\
\Rightarrow \int \frac{\mathrm{d}y}{y}=\int \frac{1}{1+x^2}\frac{1}{\arctan(x)}\, \mathrm{d}x\text{.}
$$
At this point it is helpful to note that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\arctan(x)=\frac{1}{1+x^2}\text{.}
$$
So, your integral will just be a $u$ substitution. I believe you can take it from here?

 

[spaghetti3451]

It's should be arctan(x). So
$$
y'\arctan(x)-\frac{y}{1+x^2}=0\\
\Rightarrow \int \frac{\mathrm{d}y}{y}=\int \frac{1}{1+x^2}\frac{1}{\arctan(x)}\, \mathrm{d}x\text{.}
$$
At this point it is helpful to note that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\arctan(x)=\frac{1}{1+x^2}\text{.}
$$
So, your integral will just be a $u$ substitution. I believe you can take it from here?

Yeah, I think I can do the rest. Sub u = arctan(x) and find the answer to be y = k arctan (x).

 

[SadScholar]

I believe that's correct. And you can always plug your function for y back into the differential equation and make sure that it gives you a true statement.