# Interval trouble

• Jan 25th 2011, 11:50 AM
VonNemo19
Interval trouble
Hi everyone. I'm having a lot of trouble with the idea of a solution to a differential equation HAVING to be paired with an interval in order for it to be considered a solution. So, here's a quick example direcltly from my book...

Solve $\displaystyle x\frac{dy}{dx}-4y=x^6e^x$.

Solution: Dividing by x, we get the standard form

$\displaystyle \frac{dy}{dx}-\frac{4}{x}y=x^5e^x$.

From this we identify $\displaystyle P(x)=-4/x$ and $\displaystyle f(x)=x^5e^x$ and further observe that $\displaystyle P$ and $\displaystyle f$ are continuous on $\displaystyle (0,\infty)$. Hence the integrating factor is

$\displaystyle e^{-4\int{dx/x}}=\overbrace{e^{-4\ln{x}}=e^{\ln{x^{-4}}}}^{\text{we can use } lnx\text{ instead of } ln|x| \text{ since } x>0}}=x^{-4}$...

They go on to solve the equation, but my question is why are they justified in dropping the absolute value? It is unclear to me.
• Jan 25th 2011, 12:03 PM
mr fantastic
Quote:

Originally Posted by VonNemo19
Hi everyone. I'm having a lot of trouble with the idea of a solution to a differential equation HAVING to be paired with an interval in order for it to be considered a solution. So, here's a quick example direcltly from my book...

Solve $\displaystyle x\frac{dy}{dx}-4y=x^6e^x$.

Solution: Dividing by x, we get the standard form

$\displaystyle \frac{dy}{dx}-\frac{4}{x}y=x^5e^x$.

From this we identify $\displaystyle P(x)=-4/x$ and $\displaystyle f(x)=x^5e^x$ and further observe that $\displaystyle P$ and $\displaystyle f$ are continuous on $\displaystyle (0,\infty)$. Hence the integrating factor is

$\displaystyle e^{-4\int{dx/x}}=\overbrace{e^{-4\ln{x}}=e^{\ln{x^{-4}}}}^{\text{we can use } lnx\text{ instead of } ln|x| \text{ since } x>0}}=x^{-4}$...

They go on to solve the equation, but my question is why are they justified in dropping the absolute value? It is unclear to me.

All you need is a function that allows the left hand side to be written in the form d/dx( ....) Dropping the | | does not stop the resulting function from letting you do this.
• Jan 25th 2011, 12:26 PM
dwsmith
Quote:

Originally Posted by VonNemo19
Hi everyone. I'm having a lot of trouble with the idea of a solution to a differential equation HAVING to be paired with an interval in order for it to be considered a solution. So, here's a quick example direcltly from my book...

Solve $\displaystyle x\frac{dy}{dx}-4y=x^6e^x$.

Solution: Dividing by x, we get the standard form

$\displaystyle \frac{dy}{dx}-\frac{4}{x}y=x^5e^x$.

From this we identify $\displaystyle P(x)=-4/x$ and $\displaystyle f(x)=x^5e^x$ and further observe that $\displaystyle P$ and $\displaystyle f$ are continuous on $\displaystyle (0,\infty)$. Hence the integrating factor is

$\displaystyle e^{-4\int{dx/x}}=\overbrace{e^{-4\ln{x}}=e^{\ln{x^{-4}}}}^{\text{we can use } lnx\text{ instead of } ln|x| \text{ since } x>0}}=x^{-4}$...

They go on to solve the equation, but my question is why are they justified in dropping the absolute value? It is unclear to me.

The domain of the natural log is one of the intervals P(x) and f(x) are continuous. Therefore, the natural log is restricting the x values. And since it is restricting them to its domain, why would you need the absolute value of a positive number?
• Jan 25th 2011, 01:50 PM
Ackbeet
I think, technically, the solution is valid either on $\displaystyle (-\infty,0)$ or $\displaystyle (0,\infty).$

I always keep the absolute value signs, unless I'm dealing with an integrating factor that's an even power of $\displaystyle x.$ When I'm all done solving the equation, and I'm ready to apply the initial condition, then I choose which interval I'm going to use based on which interval contains the initial condition.

Make sense?
• Jan 25th 2011, 04:48 PM
HallsofIvy
As Ackbeet pointed out, unless there is some other condition we weren't told about, we are NOT justified in just dropping the absolute value. But we can say that if x> 0, then |x|> 0 and the given solution is correct. But we could also say that if x< 0, |x|= -x and then integrating factor is x^4.

The critical point is that we have to choose one or the other- the solution cannot be continued past x= 0. Which solution is correct would depend on some additional fact- like an initial value if $\displaystyle x(t_0)> 0$, for some $\displaystyle t_0$, x will always be positive and and if $\displaystyle x(t_0)< 0$, x will always be negative.
• Jan 25th 2011, 05:32 PM
dwsmith
Quote:

Originally Posted by HallsofIvy
As Ackbeet pointed out, unless there is some other condition we weren't told about, we are NOT justified in just dropping the absolute value. But we can say that if x> 0, then |x|> 0 and the given solution is correct. But we could also say that if x< 0, |x|= -x and then integrating factor is x^4.

The critical point is that we have to choose one or the other- the solution cannot be continued past x= 0. Which solution is correct would depend on some additional fact- like an initial value if $\displaystyle x(t_0)> 0$, for some $\displaystyle t_0$, x will always be positive and and if $\displaystyle x(t_0)< 0$, x will always be negative.

I have the book he is using. When I took DE, I was confused as well, but Zills does that for some reason.