All. Looking for some help with the following integration.
If the acceleration of a particle is given as
$$ a(t) = \frac{dv(t)}{dt} = k(1+3x^2(t)) $$
where $k$ is constant (and $ \frac{dx(t)}{dt} = v(t) $), then how am I able to solve for $v(t)$?
The hint provided said to use an integrating factor but I'm not sure I'm able to put it in the required form of
$$ \frac{dy}{dx} + p(x)y(x) = q(x) $$
Or, at least, it isn't obvious to me.
Before that I tried...
$$ \frac{dv(t)}{dt} = k(1+3x^2(t)) $$
$$ \int \frac{dv(t)}{dt} dt = \int k(1+3x^2(t)) dt $$
$$ v(t) = kt + 3k\int x^2(t)dt $$
And now I'm stuck here... Any help would be appreciated; I'm hoping it's a simple mistake I can chalk up to being under caffeinated. :) Thanks!
As WW1 pointed before, make
$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v=k(1+3x^2).$$
Now, you have a separable differential equation that you can solve easily integrating
$$\int vdv=\int k(1+x^2),$$
$$\frac{v^2}{2}=k(x+x^3)+c.$$
Then,
$$v(t)=\pm \sqrt{kx(1+x^2(t))+c}$$