Question

fluid mixing question, i calculated the rate in - rate out to be 500-(y(t)/(40+t)). it's now asking me to solve the differential equation given y(0)=400grams. I know you plug that into y(t) but how do i isolate for t?

Answer 1

Plug what into where?

You have the diff.eq.
\[y'=500-\frac{y}{40+t}\]
Have you solve for \(y(t)\) yet?

Answer 2

i thought you directly plug y(0)=400 into y(t) in 500-(y(t)/(40+t))

Answer 3

The solution to the diff.eq. is \(y(t)\). First you find a general solution that contains a general constant, then use those initial conditions to solve for the constant.

Answer 4

o so you find y(t) and then plug those values in? but how do you approach dy/dx=500-(y(t)/(40+t)) when y(t) is apart of it?

Answer 5

Rearranging the equation a bit, you have
\[y'+\frac{1}{40+t}y=500\]
Linear first order ODE, so find the integrating factor:
\[\large\mu(t)=e^{\int\frac{1}{40+t}~dt}\]
Then multiply both sides by \(\mu(t)\).

Answer 6

.. i have not learned Linear first order... but ill try to multiply both sides by that and see what i get

Answer 7

Actually, all that stuff I said about the integrating factor isn't necessary. You'll find that \(\mu(t)=40+t\), which (after multiplying both sides of the equation by it) gives you
\[(40+t)y'+y=500(40+t)\]
Notice that you could have simply multiplied both sides of the equation by the denominator to get the same thing.

Answer 8

Now the trick is to notice that the left side is
\[(40+t)y'+y=\frac{d}{dt}\big[(40+t)y\big]\]
So you have
\[\frac{d}{dt}\big[(40+t)y\big]=500(40+t)\\
(40+t)y=\int500(40+t)~dt\\
~~~~~~~~~~~~~~~~~~~\vdots\]

Answer 9

so (40+t) y = 2000t+250t^2
isolate for y and plug in 400 and solve for t??

Answer 10

\[(40+t)y=2000t+250t^2+C\\
y(t)=\frac{2000t+250t^2+C}{40+t}\]
(That's the general solution above.)

Now use the initial condition: y(0)=400:
\[400=\frac{2000(0)+250(0)^2+C}{40+(0)}\]
Solve for \(C\), then put it back into the general solution.

Answer 11

oh! i see! thank you so much!!!