Verifying Differential Equations Solutions: ODEs on Intervals

[mathnerd15]

Hi! I think I have to ask this since I'm having health problems-

from Kreyszig, for xy'=-y how do you verify the solution y=h(x)=clnx by differentiating
y'=h'(x)=-clnx^2? I don't see how you get the x^2 term
also for ODEs the solution is on an open interval a<x<b but how does it include special cases of the intervals -inf<x<b, a<x<inf, -inf<x<inf; wouldn't the open interval a<x<b exclude -inf<x<b?
thanks very much!

 

[lurflurf]

That is not a solution.
It is a separable differential equation.

 

[mathnerd15]

thanks very much!
I separated leading to-
-ln|y|=ln|x|+C
so then you just find the substitution c/x? (I misread the text: l is /)

 

[HallsofIvy]

thanks very much!
I separated leading to-
-ln|y|=ln|x|+C
so then you just find the substitution c/x? (I misread the text: l is /)

Well, you don't just "find" a substitution. You always solve an equation of the form f(y)= F(x) for y by taking $f^{-1}$ of both sides. The inverse function to ln(x) is, of course, $e^x$ so take the exponential of both sides:
$$e^{-ln|y|}= e^{ln|x|+ C}$$

$-ln|y|= ln|y^{-1}|$ so the left side is $y^{-1}= 1/y$. Because $e^{a+ b}= e^ae^b$ the right side is $e^{ln|x|}e^C= C' |x|$ where C'= e^C.
That is, $1/|y|= C'|x|$ or $|y|= C'/|x|$. We can then "absorb" the absolute values into C' by allowing it to be positive or negative.

 

[blue_raver22]

I would like to clarify that verifying solutions to differential equations is an important step in the process of solving ODEs. In order to verify the solution y=h(x)=clnx by differentiating, we can use the chain rule and product rule to differentiate both sides of the equation. This will result in y'=h'(x)=-clnx^2, as correctly stated in the question. The x^2 term comes from differentiating the ln function using the power rule.

Regarding the intervals for ODE solutions, it is important to note that the open interval a<x<b includes all values between a and b, including special cases such as -inf<x<b, a<x<inf, and -inf<x<inf. This is because an open interval does not have a specific endpoint, so it encompasses all values within its boundary. Therefore, the open interval a<x<b does not exclude any of the mentioned special cases.

I hope this clarifies any confusion regarding verifying ODE solutions and the use of open intervals. If you are experiencing health problems, I would recommend seeking professional medical advice. I am not qualified to provide medical advice.