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Math Help - How to treat dy/dx when integrating

  1. #1
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    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 )
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  2. #2
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    Re: How to treat dy/dx when integrating

    You should turn y' into dy/dx to separate variables.

    \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

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

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

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

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

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

    Now you can integrate both sides
    Thanks from Wolvenmoon and HallsofIvy
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