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We have the following ODE: $ \dfrac{dy}{dx} = \dfrac{x}{y}$

If we want to solve it we just rewrite this to the form: $ydy = xdx$ and we then take the integral, but here is where I get confused. My textbook says this is the integral:

$\dfrac{y^2}{2} + C_1 = \dfrac{x^2}{2} + C_2 $, but isn't the integral of dx just x and the integral of dy just y? Where are these left out?

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Maybe this is clearer than the "differentials" manipulation. You can write your equation as

$$y'=\frac x y$$ or $$y'y=x$$

This is really


Upon integration we get that

$$\frac 1 2 y(x)^2=\frac{x^2}2+C$$

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I'm sorry, I haven't done integrals in a while, so I'm rusty. But why isn't the integral of $y'y$ equal to $\dfrac{1}{2}y^2 \times y = \dfrac{1}{2}y^3 $? – Ylyk Coitus Feb 21 '13 at 19:58
Take the derivative with respect to $x$ of $\frac 1 2 y(x)^3$. What do you get? Is it $y(x)\times y'(x)$? Take the derivative of $\frac 1 2 y(x)^2$ with respct to $x$. What do you get? Remember the chain rule. Being rusty with these integrals is equivalent to being rusty with derivatives. – Pedro Tamaroff Feb 21 '13 at 20:03
Yes, I see I made a mistake, but I don't see where exactly the $y'$ goes when integrated. Because I know the derivative of $\dfrac{1}{2}y^2$ is just $y$. – Ylyk Coitus Feb 21 '13 at 20:08
@YlykCoitus The derivative of $\frac{1}{2}y^{2}$ with respect to $y$ is $y \frac{dy}{dx}=yy'$. Remember the chain rule. – Daniel Littlewood Feb 21 '13 at 20:13
@YlykCoitus Sorry, that's a typo. It should say 'with respect to $x$' – Daniel Littlewood Feb 21 '13 at 20:23

Yes, but you are integrating $y\ dy$ and $x\ dx$, not $dy$ and $dx$.

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the integral of $ x \ \ dx$ is $\frac{1}{2} x^2+c$

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