Differential Equations: Separable Equations: dy/dt=2y+1

lbros55

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[FONT=&quot]the problem is: dy/dt=2y+1[/FONT]
[FONT=&quot]
[/FONT]

[FONT=&quot]I got y=1/2(e2(t+c) -1) while the book solution is y=ke2t-1/2. Would my solution be acceptable? I also have to solve the ivp for this problem which is y(0)=3. Using my first solution I first did:[/FONT]
[FONT=&quot]3=1/2(eo *e2c)-1[/FONT]
[FONT=&quot]4=1/2(e2c)[/FONT]
[FONT=&quot]8=e2c[/FONT]
[FONT=&quot]How do I solve for c or did I do this wrong?[/FONT]
 
the problem is: dy/dt=2y+1


I got y=1/2(e2(t+c) -1) while the book solution is y=ke2t-1/2. Would my solution be acceptable? I also have to solve the ivp for this problem which is y(0)=3. Using my first solution I first did:
3=1/2(eo *e2c)-1
4=1/2(e2c)
8=e2c
How do I solve for c or did I do this wrong?
You and the book agree. If you compare the two solutions you will see that \(\displaystyle k = \frac{1}{2} e^{2c}\).

So, you need to solve \(\displaystyle 8 = e^{2c}\) for c. How about you take the natural log of both sides?
\(\displaystyle ln(8) = ln \left ( e^{2c} \right ) = 2c\)

Now solve for c.

-Dan

Addendum: I should mention that I prefer the form \(\displaystyle e^{2t + 2c}\). Generally we call both k and c arbitrary constants... meaning they can be positive, negative, or zero. But if you take a look at \(\displaystyle k = e^{2c}\) you will note that k can neither never be negative or zero! c on the other hand can be anything we want it to be.
 
Last edited:
You and the book agree. If you compare the two solutions you will see that \(\displaystyle k = \frac{1}{2} e^{2c}\).

So, you need to solve \(\displaystyle 8 = e^{2c}\) for c. How about you take the natural log of both sides?
\(\displaystyle ln(8) = ln \left ( e^{2c} \right ) = 2c\)

Now solve for c.

-Dan

Addendum: I should mention that I prefer the form \(\displaystyle e^{2t + 2c}\). Generally we call both k and c arbitrary constants... meaning they can be positive, negative, or zero. But if you take a look at \(\displaystyle k = e^{2c}\) you will note that k can neither never be negative or zero! c on the other hand can be anything we want it to be.

It looks like you tried to use a superscript, but slipped. You meant to say y=1/2(e2(t+c) -1). Without formatting, this could also be written as y=1/2(e^{2(t+c)} -1).

Your answer is correct; they have gone further in simplifying before finding the constant. (This should not be necessary, so your answer should get full credit for the part you did). Here is what they did to get from your answer to theirs:

\(\displaystyle y = \frac{1}{2}\left(e^{2(t+c)} -1\right) = \frac{1}{2}\left(e^{2c}e^{2t} -1\right) = \left(\frac{e^{2c}}{2}\right)e^{2t} -\frac{1}{2} = k e^{2t} -\frac{1}{2}\),

because \(\displaystyle \left(\frac{e^{2c}}{2}\right)\) is a constant they choose to call k.

From here, they solve for k; you can solve for c as you've been shown. The result will be the same answer.

Incidentally, when you integrated you apparently neglected to write ln|2y + 1|; if you did, you would end up with a ± in your answer, as a result of which it turns out the k can be either positive or negative (but not zero), so their form is actually more correct than yours.
 
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