# Thread: Can't integrate this differential equation

1. ## [SOLVED] Integration of D.E.

Hi, having difficulty with this D.E. that relates to a tank of initially 400L of pure water, with 1L of salt water (concentration 4g/L) added per min, and 3L of the well-mixed tank solution removed per min. This is what I have so far;

$Q'(t)+\frac{3Q(t)}{400-2t}=4$

I'm trying to find an integrating factor though I keep ending up with a final differentiated x(t) function that doesn't make sense (based on t-200, resulting in the function only existing from t > 200 when it should be from t > 0)

$P(t)=\frac{3}{2(200-t)}$ leading to

$I(t)=-(200-t)^{3/2}$

but then this doesn't seem to make the LHS equal to the derivative of $Q(t)I(t)$? Without the negative sign it would work fine but I'm sure that's the integral of P(t)... a little help? Maybe I've done something stupid...

2. Hello, McChickenb!

$\frac{dQ}{dt} +\frac{3}{400-2t}\,Q\;=\;4$

We have: . $\frac{dQ}{dt} + \frac{3}{2}\cdot\frac{1}{200-t}\,Q \;=\;4$

Integrating factor: . $I \;=\;e^{\frac{3}{2}\!\int \!\frac{dt}{200-t}} \;=\; e^{-\frac{3}{2}\ln(200-t)} \;=\;e^{\ln(200-t)^{-\frac{3}{2}}} \;=\;(200-t)^{-\frac{3}{2}}$

Multiply by $I\!:\;\;(200-t)^{-\frac{3}{2}}\,\frac{dQ}{dt} + \frac{3}{2}\!\cdot\!\frac{1}{200-t}\!\cdot\!(200-t)^{-\frac{3}{2}} \,Q \;=\;4(200-t)^{-\frac{3}{2}}$

. . . . . . . . . . . . . . . . . . $(200-t)^{-\frac{3}{2}}+ \frac{3}{2}\,(200-t)^{-\frac{5}{2}}\,Q \;=\;4(200-t)^{-\frac{3}{2}}$

We have: . $\frac{d}{dt}\bigg[(200-t)^{-\frac{3}{2}}\,Q\bigg] \;=\;4(200-t)^{-\frac{3}{2}}$

Integrate: . $(200-t)^{-\frac{3}{2}}\,Q \;=\;8(200-t)^{-\frac{1}{2}} + C$

Multiply by $(200-t)^{\frac{3}{2}}\!:\;\;\;Q \;=\;8(200-t) + C(200-t)^{\frac{3}{2}}$

3. D'oh! That's perfect. I was confusing myself on the simple stuff. Thanks a lot!