Using an integration factor to integrate how would you integrate:
$$\dfrac{dx}{dt}+\dfrac{2x}{50+t}=6.$$
I've gotten the integrating factor, which is $50+t$, but where do I go from there?
Thank You.
 A: Your differential equation is of the form 
$$\frac{dx}{dt}+p(t)x = 6.$$ 
The integrating factor is then $q(t) = \exp(\int p(t) dt)$ (check this as what you've written above is not the correct integrating factor, I'm guessing because you missed a constant in $p(t)$). Once you have $q(t)$, multiply both sides of the equation by it to obtain 
$$q(t)\frac{dx}{dt} + p(t)q(t)x = 6q(t).$$
Now note that $\frac{dq}{dt} = p(t)q(t)$ (by design), so we have
$$q(t)\frac{dx}{dt}+\frac{dq}{dt}x = 6q(t)$$
which can be written as
$$\frac{d}{dt}(q(t)x) = 6q(t)$$
by the product rule. Therefore, $q(t)x = \int 6q(t)dt$ so
$$x = \frac{1}{q(t)}\int 6q(t) dt.$$
A: Multiply by $e^\int{\frac{2}{50+t}}$ on both sides. The LHS becomes a perfect differential. Integrate on both sides to get
$$ x (50+t)^2  =  6\int{(50+t)^2}$$
Can you carry on from there?
A: If the first order linear DE
$$\frac{dx}{dt}+P(t)x=Q(t)$$
has integrating factor $I(t)$ then multiplying both sides by $I(t)$ gives
$$I(t)\frac{dx}{dt}+I(t)P(t)x=I(t)Q(t)\ .$$
This can be written
$$\frac{d}{dt}\bigl(I(t)x\bigr)=I(t)Q(t)$$
which is now easy to integrate.  At least, the LHS is ;-)
A: Your integrating factor is wrong. It should be $e^{2\ln(50+t)} = (50+t)^2$.
Multiplying throughout by that you get:
$(50+t)^2\frac{dx}{dt} + 2x(50+t) = 6(50+t)^2$
$\frac{d}{dt}[x(50+t)^2] = 6(50+t)^2$
It now becomes a separable d.e. you can easily solve.
