# Find the exact value of a differential equation.

1. Oct 19, 2016

### Sam Donovan

1. The problem statement, all variables and given/known data
dy/dx= 200-2y. y(0)=75

2. Relevant equations

3. The attempt at a solution
Do you move dx over and integrate.

Do you just integrate it 200y-y^2+c

2. Oct 19, 2016

### andrewkirk

3. Oct 19, 2016

### Sam Donovan

I'm having a little trouble with this. I move it so it becomes

$dy/dx=200-2y$
$dy=200dx-2ydx$

4. Oct 19, 2016

### Staff: Mentor

No. This is completely wrong. I would strongly advise you to look at some examples of this technique in your textbook.

Using separation of variables, you should end up with all terms involving y and dy on one side, and all terms involving x and dx on the other side.
For this problem,
1) Divide both sides by 200 - 2y
2) Multiply both sides of the resulting equation by dx

In this case, you should end up with $\frac{dy}{200 - 2y} = dx$, or equivalently, $\frac{dy}{100 - y} = 2 dx$,

5. Oct 21, 2016

### vela

Staff Emeritus
I think what you're suggesting is to solve the problem as follows:
\begin{align*}
\frac{dy}{dx} &= 200-2y \\
\int \frac{dy}{dx}\,dx &= \int (200-2y)\,dx \\
y &= 200y-y^2+c
\end{align*}
There's a problem with that last step. If you don't recognize it, consider the questions below:

Outside of this problem, if I asked you what $\int 200\,dx$ equalled, you'd hopefully say $200 x+c$, yet in solving this problem, you encountered the exact same integral and (allegedly) said $\int 200\,dx = 200y+c$.

Similarly, consider the two integrals $\int 2x\,dx$ and $\int 2y\,dx$. Why can you easily evaluate the first one but not the latter?