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Solution for ∫y(x)dx = ky, = k/y, = kx, = k/x

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Get the solution (for y(x)) for the follows integrals: [tex]\int y(x) dx = ky\;\;\;\;\;(1)[/tex] [tex]\int y(x) dx = \frac{k}{y}\;\;\;\;\;(2)[/tex] [tex]\int y(x) dx = kx\;\;\;\;\;(3)[/tex] [tex]\int y(x) dx = \frac{k}{x}\;\;\;\;\;(4)[/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex]\\ \int y(x)dx = kx \\ \\ \int y(x)\frac{dx}{dx} = \frac{kx}{dx} \\ \\ d\int y(x) = d\frac{kx}{dx} \\ \\ y(x) = k\frac{dx}{dx} \\ \\ y(x) = k[/tex]

    [tex]\\ \int y(x)dx = \frac{k}{x} \\ \\ \int y(x)\frac{dx}{dx} = \frac{1}{dx} \frac{k}{x} \\ \\ d\int y(x) = \frac{d}{dx} \frac{k}{x} \\ \\ y(x) = k \frac{d}{dx}\left ( \frac{1}{x} \right ) \\ \\ y(x) = -\frac{k}{x^2}[/tex]
     
  2. jcsd
  3. Mar 8, 2014 #2

    Mark44

    Staff: Mentor

    I'm not sure anyone would buy your reasoning, especially dividing by dx that you have done.
    For problem 3, ∫y(x)dx = kx, differentiate both sides with respect to x to get y(x) = k.
    For problem 4, ∫y(x)dx = k/x, differentiate both sides with respect to x to get y(x) = -k/x2.
    For problem 1, which you didn't try, ∫y(x)dx = ky. As before, differentiate both sides with respect to x to get y(x) = k dy/dx. This is a separable differential equation that is pretty easy to solve.
     
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