Integrating y * dy/dt = x * dx/dt

**algorithm** · Oct 16th 2012, 07:30 AM

Hi,

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

My textbook says the integral is:

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

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

Thanks.

**TheEmptySet** · Oct 16th 2012, 08:04 AM

Originally Posted by algorithm

Hi,

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

My textbook says the integral is:

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

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

Thanks.

My guess is that they are not.

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

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

$ydy=xdx$

Now integrate both sides to get

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

**JJacquelin** · Oct 16th 2012, 08:14 AM

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

**algorithm** · Oct 16th 2012, 11:42 AM

I see. Thanks for your explanations.

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