# How to treat dy/dx when integrating

• Aug 28th 2013, 02:53 PM
Wolvenmoon
How to treat dy/dx when integrating
Hello all!

I'm feeling pretty sheepish. I finished calc 3 two years ago and am just starting diff EQ. The first section is kicking my butt pretty hard.

Here's my current homework problem:

y'=(x-3)(y^2+1), y(0)=1

Here's what I've done so far:

y'/(y^2+1)=(x-3)

Integral of (y'/(y^2+1)) = integral of (x-3)

arctan(y)=(x^2/2-3x+C)

My issue is in how I treat Y prime when integrating. I know the answer I have is correct so far - Wolfram Alpha says so - but I've lost why y' (AKA dy/dx) is being treated as "1". I understand that it is, I've lost why it is.

My book is concise to a fault, so I need some help understanding this because it's a basic building block that I should remember but don't. I forgot that integration as well - I had to look up what it took to get an arctan after integrating!

(Answer is: arctan(1) = 0-0+C, C= pi/4
arctan(y) = x^2/2 -3x + pi/4

Book simplifies it to: arctan(y) = (2x^2-12x+pi)/4 )
• Aug 28th 2013, 03:03 PM
Shakarri
Re: How to treat dy/dx when integrating
You should turn y' into dy/dx to separate variables.

$\displaystyle \int \frac{y'}{y^2+1}=\int x-3$
From here you cannot integrate the right hand side since there is no dx in the integral.
Instead you should treat the problem like this

$\displaystyle \frac{y'}{y^2+1}= x-3$

$\displaystyle y' \frac{1}{y^2+1}= x-3$

$\displaystyle \frac{dy}{dx} \frac{1}{y^2+1}= x-3$

$\displaystyle \frac{dy}{y^2+1}= (x-3) dx$

$\displaystyle \int \frac{dy}{y^2+1}= \int (x-3) dx$

Now you can integrate both sides