First-Order Separable ODE

**joatmon** · September 29th 2011, 07:35 PM

I feel like I've been thrown into a pool with no swimming lessons...

Thanks for all the help on these tonight. I have quite a few more, and very little instruction as to how to address these, so I'm sure there will be more posts.

Here's the latest one:

Find the solution of the differential equation that satisfies the given initial condition.

Cutting through the algebra...

$\frac{dP}{\sqrt{Pt}} = 2 dt$

$\frac{1}{t} \int \frac{du}{\sqrt{u}} = 2 \int dt$

$\frac{2}{t}\sqrt{Pt} = 2t + C$

Let me just stop here for the moment. If I multiply both sided by t/2, can I leave the C as is, or does this also have to be multiplied by t/2? This is tricky - knowing when to leave the C alone or when to include it with the algebra.

**Soroban** · September 29th 2011, 07:53 PM

Hello, joatmon!

Find the solution of the differential equation that satisfies the given initial condition.

. . $\frac{dP}{dt} \:=\:2\sqrt{Pt},\;\;P(1) = 7$

We have: . $\frac{dP}{dt} \;=\;2\sqrt{P}\,\sqrt{t} \quad\Rightarrow\quad \frac{dP}{\sqrt{P}} \;=\;2\sqrt{t}\,dt$

$\text{Integrate: }\:\int P^{-\frac{1}{2}}\,dP \;=\;2\int t^{\frac{1}{2}}\,dt \quad\Rightarrow\quad 2P^{\frac{1}{2}} \;=\;\tfrac{4}{3}t^{\frac{3}{2}} + C$

Since $P(1) = 7\!:\;\;2(7^{\frac{1}{2}}) \;=\;\tfrac{4}{3}(1^{\frac{3}{2}}) + C \quad\Rightarrow\quad C \;=\;2\sqrt{7} - \tfrac{4}{3}$

Therefore: . $2\sqrt{P} \;=\;\tfrac{4}{3}\sqrt{t^3} + 2\sqrt{7} - \tfrac{4}{3}$

**joatmon** · September 29th 2011, 08:10 PM

Thanks. You guys always make this stuff look so easy and obvious...

In this case, though, I think that there is more work to be done. After you integrated, don't you have to continue on to solve for P before solving for C?

$P^{\frac{1}{2}} = \frac{2}{3}t^{\frac{3}{2}} + C$

$P = \frac{4}{9}t^3 + C$

Then, using the initial value information given:
$7 = \frac{4}{9}(1)^3 + C$

$C = \frac{59}{9}$

Therefore: $P = \frac{4}{9}t^3 + \frac{59}{9}$

This is a different answer than you got. Since I have to solve for P, do you agree with what I did? Thanks!

**joatmon** · September 29th 2011, 09:11 PM

I entered this answer, but it didn't check out. Does anyone see what I am doing wrong? Thanks.

**mr fantastic** · September 29th 2011, 10:36 PM

Originally Posted by joatmon

I entered this answer, but it didn't check out. Does anyone see what I am doing wrong? Thanks.

Post #2 told you how to get $\sqrt{P}$ . To get P, you have to - correctly - square both sides.

**joatmon** · September 30th 2011, 06:08 AM

Which I tried to do, showing you the answer that I came up with, and that I knew that it was wrong.

**Ackbeet** · September 30th 2011, 06:31 AM

Originally Posted by joatmon

Thanks. You guys always make this stuff look so easy and obvious...

In this case, though, I think that there is more work to be done. After you integrated, don't you have to continue on to solve for P before solving for C?

$P^{\frac{1}{2}} = \frac{2}{3}t^{\frac{3}{2}} + C$

$P = \frac{4}{9}t^3 + C$

This is what mr f was getting at. When you squared the previous equation to get this one, you didn't square correctly.

Then, using the initial value information given:
$7 = \frac{4}{9}(1)^3 + C$

$C = \frac{59}{9}$

Therefore: $P = \frac{4}{9}t^3 + \frac{59}{9}$

This is a different answer than you got. Since I have to solve for P, do you agree with what I did? Thanks!

**joatmon** · September 30th 2011, 09:47 AM

I have verified that $(\frac{2}{3}t^{\frac{3}{2}})^2 = \frac{4}{9}t^3$ , so I'm looking for a little more of a suggestion. I tried including the C in the squaring, but this just gave me a crazy answer that also didn't work out correctly. For sure, I might be doing something wrong with that calculation, but my basic source of confusion at this point is how I should "correctly" square the right side of this equation.

Please have a little mercy here. This is my first exposure to differential equations. So far, in all of our integrals, we just leave the C alone, but I think that in this case, I have to do something with it, but I don't know exactly what that is. The bigger question is "when do I leave the C alone and when do I wrap it up in the rest of the algebra?". Are there any rules for this? Our text book doesn't get into this at all, but here it seems to be elemental to solving this.

Thanks for helping me. I'm frustrated, but I appreciate your help.

**Ackbeet** · September 30th 2011, 09:53 AM

Originally Posted by joatmon

Thanks. You guys always make this stuff look so easy and obvious...

In this case, though, I think that there is more work to be done. After you integrated, don't you have to continue on to solve for P before solving for C?

$P^{\frac{1}{2}} = \frac{2}{3}t^{\frac{3}{2}} + C$

$P = \frac{4}{9}t^3 + C$

You should have gotten

$P=\frac{4}{9}\,t^{3}+\frac{4C}{3}\,t^{3/2}+C^{2}.$

The rule concerning constants, and when you can absorb them is this: when a new expression results containing all constants, sometimes you can simply re-define the constant to include any other constants that affect the original constant via basic four arithmetic or exponentiation. However, in the example above, you cannot redefined the middle term to be an arbitrary constant, because it has a t in it.

Does that help?

Then, using the initial value information given:
$7 = \frac{4}{9}(1)^3 + C$

$C = \frac{59}{9}$

Therefore: $P = \frac{4}{9}t^3 + \frac{59}{9}$

This is a different answer than you got. Since I have to solve for P, do you agree with what I did? Thanks!

**mr fantastic** · September 30th 2011, 02:18 PM

Originally Posted by joatmon

I have verified that $(\frac{2}{3}t^{\frac{3}{2}})^2 = \frac{4}{9}t^3$ , so I'm looking for a little more of a suggestion. I tried including the C in the squaring, but this just gave me a crazy answer that also didn't work out correctly. For sure, I might be doing something wrong with that calculation, but my basic source of confusion at this point is how I should "correctly" square the right side of this equation.

Please have a little mercy here. This is my first exposure to differential equations. So far, in all of our integrals, we just leave the C alone, but I think that in this case, I have to do something with it, but I don't know exactly what that is. The bigger question is "when do I leave the C alone and when do I wrap it up in the rest of the algebra?". Are there any rules for this? Our text book doesn't get into this at all, but here it seems to be elemental to solving this.

Thanks for helping me. I'm frustrated, but I appreciate your help.

It is not an issue of "first exposure to differential equations', it is an issue of basic algebra. In particluar, how to square an expression.

If you showed all the algebra of your working, the exact error you're making could have been pointed out with surgical precision.

Note: When studying differential equations, it is assumed that a student has mastery of algebra and basic calculus techniques.

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