1. Dif EQ

Another Dif EQ, yay.

I have to solve the following Dif EQ, and give the interval for which the largest general solution is defined:

dP/dt + 2tP = P + 4t - 2

So I don't think I can do separation of variables, so I tried integrating factors...

P(x) = 2t

M(x) = e^(int(2t)dt) = e^(2t^3/3)

So multiply M(x) to each term:

e^(2t^3/3)*dP/dt + e^(2t^3/3)*2tP = e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3)

So this looks very messy now...

Again, it's the part after this that I get stuck.

2. Upon re-examining this,

Would it be:

int((e^(2t^3/3)*P)') = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

=> (e^(2t^3/3)*P) = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

3. Originally Posted by Ideasman
Another Dif EQ, yay.

I have to solve the following Dif EQ, and give the interval for which the largest general solution is defined:

dP/dt + 2tP = P + 4t - 2

So I don't think I can do separation of variables, so I tried integrating factors...

P(x) = 2t
you are wrong since this line

Originally Posted by Ideasman
Upon re-examining this,

Would it be:

int((e^(2t^3/3)*P)') = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

=> (e^(2t^3/3)*P) = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))
thus, this is wrong as well

4. Originally Posted by Jhevon
you are wrong since this line
But why? I thought you took the stuff that wasn't on the dependent variable and let that equal P(x). Is it because of the RHS of the equation? Do I need to move things first? Suppose it was just

dP/dt + 2tP = 0. Would P(x) be 2t then?

5. Originally Posted by Ideasman
But why? I thought you took the stuff that wasn't on the dependent variable and let that equal P(x). Is it because of the RHS of the equation? Do I need to move things first?
you had $\displaystyle P' + 2tP = P + 4t - 2$

$\displaystyle \Rightarrow P' + 2tP - P = 4t - 2$

$\displaystyle \Rightarrow P' + (2t - 1)P = 4t - 2$

now what do you think $\displaystyle P(x)$ is?

(in this case, using "P" gets confusing, because that's the variable we're working with in the first place, but oh well)

Suppose it was just

dP/dt + 2tP = 0. Would P(x) be 2t then?
yes, but you would not use the integrating factor in this case, this is a homogeneous first order DE, it is separable

6. Originally Posted by Jhevon
you had $\displaystyle P' + 2tP = P + 4t - 2$

$\displaystyle \Rightarrow P' + 2tP - P = 4t - 2$

$\displaystyle \Rightarrow P' + (2t - 1)P = 4t - 2$

now what do you think $\displaystyle P(x)$ is?
2t - 1

Okay, so if P(x) = 2t - 1

That means M(x) = e^(int(2t - 1)dt) = e^(t^2 - t)

Multiply M(x) by each term:

e^(t^2 - t)*P' + e^(t^2 - t)*P = e^(t^2 - t)*4t - 2*e^(t^2 - t)

Okay:

int[e^(t^2 - t)*P)'] = int(e^(t^2 - t)*4t - 2*e^(t^2 - t))

=> e^(t^2 - t)*P = {Since this can't be integrated, I think I've done something wrong }

7. Originally Posted by Ideasman
2t - 1

Okay, so if P(x) = 2t - 1

That means M(x) = e^(int(2t - 1)dt) = e^(t^2 - t)

Multiply M(x) by each term:

e^(t^2 - t)*P' + e^(t^2 - t)*P = e^(t^2 - t)*4t - 2*e^(t^2 - t)

Okay:

int[e^(t^2 - t)*P)'] = int(e^(t^2 - t)*4t - 2*e^(t^2 - t))

=> e^(t^2 - t)*P = {Since this can't be integrated, I think I've done something wrong }
no, you're correct. new strategy: let's go for separable

$\displaystyle P' + (2t - 1)P = 4t - 2$

$\displaystyle \Rightarrow P' = 4t - 2 - (2t - 1)P$

$\displaystyle \Rightarrow P' = 2(2t - 1) - (2t - 1)P$

$\displaystyle \Rightarrow P' = (2t - 1)(2 - P)$

$\displaystyle \Rightarrow \frac {P'}{2 - P} = 2t - 1$

now continue

8. Originally Posted by Jhevon
no, you're correct. new strategy: let's go for separable

$\displaystyle P' + (2t - 1)P = 4t - 2$

$\displaystyle \Rightarrow P' = 4t - 2 - (2t - 1)P$

$\displaystyle \Rightarrow P' = 2(2t - 1) - (2t - 1)P$

$\displaystyle \Rightarrow P' = (2t - 1)(2 - P)$

$\displaystyle \Rightarrow \frac {P'}{2 - P} = 2t - 1$

now continue
Ah, tricky algebra manipulation.

Okay, so from where you left off:

dP/dt * 1/(2-P) = 2t - 1

=> dP/(2-P) = (2t-1)dt

Take integral of both sides:

-ln|P-2| = t^2 - t + C

Now the second part of the question asked to find the largest interval over which the GENERAL sol'n is defined. This looks defined everywhere, except for when the ln is 0 or negative, but since we have the absolute signs, we don't have to worry about it being neg. Therefore, it's defined (-infinity, 2) or (2, infinity). How am I supposed to know which it's in though.

9. Originally Posted by Ideasman
Ah, tricky algebra manipulation.

Okay, so from where you left off:

dP/dt * 1/(2-P) = 2t - 1

=> dP/(2-P) = (2t-1)dt

Take integral of both sides:

-ln|P-2| = t^2 - t + C

Now the second part of the question asked to find the largest interval over which the GENERAL sol'n is defined. This looks defined everywhere, except for when the ln is 0 or negative, but since we have the absolute signs, we don't have to worry about it being neg. Therefore, it's defined (-infinity, 2) or (2, infinity). How am I supposed to know which it's in though.
read, again, what i said here regarding this question

10. Originally Posted by Jhevon
read, again, what i said here regarding this question
That's the whole point; I don't have a point t_0.

11. Originally Posted by Ideasman
That's the whole point; I don't have a point t_0.
what is P(x)? and where is it continuous?