Separable DE Mixing problem

[Cookiemnstr510]

Hello All,
I previously took DE and I am trying to brush up a bit. I am stuck on a mixing problem right now. I will post the problem below. I am wondering how they get the answer in the book. It seems like they skip multiple steps and I would like to see those steps.
I have also attached my attempt at the problem.THE PHOTOS ARE A BIT OUT OF ORDER, the last one is the first one, the. The rest are in order.
Thanks

 

[MarkFL]

Hello, and welcome to FMH!

Your book gives the IVP:

[MATH]\d{x}{t}=\frac{3}{5}-\frac{3}{500}x[/MATH] where \(x(0)=0\)

Instead of treating the ODE as separable, I would treat is as a linear equation, and write it in standard form:

[MATH]\d{x}{t}+\frac{3}{500}x=\frac{3}{5}[/MATH]
Compute the integrating factor \(\mu(t)\):

[MATH]\mu(t)=\exp\left(\frac{3}{500}\int\,dt\right)=e^{\frac{3}{500}t}[/MATH]
And so the ODE becomes:

[MATH]e^{\frac{3}{500}t}\d{x}{t}+\frac{3}{500}e^{\frac{3}{500}t}x=\frac{3}{5}e^{\frac{3}{500}t}[/MATH]
Rewrite the LHS:

[MATH]\frac{d}{dt}\left(e^{\frac{3}{500}t}x\right)=\frac{3}{5}e^{\frac{3}{500}t}[/MATH]
Integrate:

[MATH]\int\frac{d}{dt}\left(e^{\frac{3}{500}t}x\right)\,dt=\frac{3}{5}\int e^{\frac{3}{500}t}\,dt[/MATH]
[MATH]e^{\frac{3}{500}t}x=100e^{\frac{3}{500}t}+c_1[/MATH]
[MATH]x(t)=100+c_1e^{-\frac{3}{500}t}[/MATH]
Using the initial values to determine the value of the parameter:

[MATH]x(0)=100+c_1=0\implies c_1=-100[/MATH]
And so the solution to the IVP is:

[MATH]x(t)=100-100e^{-\frac{3}{500}t}=100\left(1-e^{-\frac{3}{500}t}\right)[/MATH]

 

[HallsofIvy]

You can, of course, do this as a "separable equation", \(\displaystyle \frac{dx}{\frac{3}{5}-\frac{3}{500}x}= dt\) and integrate. To integrate on the left I would first multiply numerator and denominator by 500 to eliminate the fractions, so \(\displaystyle \frac{500dx}{300- 3x}= dt\). Now let y= 300- 3x so that dy= -3dx and \(\displaystyle dx= -\frac{dy}{3}\). The equation becomes \(\displaystyle -\frac{500}{3}\frac{dy}{y}= dt\) or \(\displaystyle \frac{dy}{y}= -\frac{3}{500}dt\).

Integrating, \(\displaystyle ln(y)= -\frac{3}{500}t+ C\). Taking the exponential of both sides, \(\displaystyle y= C' e^{-\frac{3t}{500}}\) where \(\displaystyle C'= e^C\). Replacing y by 300- 3x, \(\displaystyle 300- 3x= C' e^{-\frac{3t}{500}}\) so \(\displaystyle x= 100- \frac{C'}{3}e^{-\frac{3t}{500}}\). That is exactly what MarkFL got, except that his \(\displaystyle c_1\) is the negative of my C'/3. (And, of course, it is much more tedious!)