# Thread: Integrating y * dy/dt = x * dx/dt

1. ## Integrating y * dy/dt = x * dx/dt

Hi,

$\displaystyle y(t) * dy(t)/dt = x(t) * dx(t)/dt$

My textbook says the integral is:

$\displaystyle y^2(t) - x^2(t) = constant$.

The part I'm not sure about is how they can put $\displaystyle y(t)$ out of the integral sign, and just integrate $\displaystyle dy(t)/dt$, as $\displaystyle y(t)$ is a function of time.

Thanks.

2. ## Re: Integrating y * dy/dt = x * dx/dt

Originally Posted by algorithm
Hi,

$\displaystyle y(t) * dy(t)/dt = x(t) * dx(t)/dt$

My textbook says the integral is:

$\displaystyle y^2(t) - x^2(t) = constant$.

The part I'm not sure about is how they can put $\displaystyle y(t)$ out of the integral sign, and just integrate $\displaystyle dy(t)/dt$, as $\displaystyle y(t)$ is a function of time.

Thanks.
My guess is that they are not.

$\displaystyle y\frac{dy}{dt}=x\frac{dx}{dt}$

With a little abuse of notation multiply both sides of the equation by $\displaystyle dt$

$\displaystyle ydy=xdx$

Now integrate both sides to get

$\displaystyle \int ydy = \int xdx \implies \frac{1}{2}y^2=\frac{1}{2}x^2+c \iff y^2-x^2=\hat{c}$

3. ## Re: Integrating y * dy/dt = x * dx/dt

y dy/dt = (1/2) d(y²)/dt
x dx/dt = (1/2) d(x²)/dt
y dy/dt = x dx/dt hence d(y²)/dt = d(x²)/dt
Since the derivatives are equal, the functions differ only from a constant;
y²-x² = c

4. ## Re: Integrating y * dy/dt = x * dx/dt

I see. Thanks for your explanations.

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