Verify that the indicated function is a solution to the given differential equation

$\displaystyle xy'-sin(x)=0$, *$\displaystyle y = \int_{0}^{x}\frac{sin(t)}{T}dt$*

So this is what I did. I integrated the function to get, $\displaystyle \frac{1}{T}[-cos(x)+cos(1)]$

Then took the derivative of that with respect to x to get $\displaystyle \frac{sin(x)}{T}$

Then I sub it in to get $\displaystyle \frac{xsin(x)}{T} - sin(x)$ which does not equal 0, so i'm saying that this function is not a solution to the differential equation.

IS this correct?

David.

Re: Verify that the indicated function is a solution to the given differential equati

I see no reason to integrate when you just turn around and differentiate. With $\displaystyle y= \int_0^x\frac{sin(t)}{T}dt$ by the "Fundamental Theorem of Calculus", that the derivative of $\displaystyle \int_a^x f(t) dx$ is just f(x), we get immediately your $\displaystyle y'= \frac{sin(t)}{T}$. And that does NOT satisfy the differential equation, again as you say.

HOWEVER, I think you have misread the problem. If $\displaystyle y= \int_0^x \frac{sin(t)}{t}dt$, so that the denominator is the variable, not a constant, then the function becomes very hard to integrate but it is easy to use the "Fundamental Theorem of Calculus" to get $\displaystyle y'= \frac{sin(x)}{x}$. And that clearly **does** satisfy the equation.

(By the way, the constant in your integral should be cos(0)= 1, not cos(0)- but that was probably a typo.)

Re: Verify that the indicated function is a solution to the given differential equati

Yeah sorry about that. The typo was actually in the question, not my answer. The limits of integration was from 1 - x. So it should have been a 1 at the bottom end of the integral sign. ANd the question had a capital T for the denominator as I wrote. But I think you're right and actually the capital T should have been a small t, because they give a hint using the fundamental theorum of calculus just like you stated when you were talking about it should have been a small t.

Yeah, so I will assume they meant a small t instead of a big T and use the Fundamental THeorem of Calculus as you did.

Thanks for your help. I don't know if the 1 makes that harder now, but maybe that was a typo in the book too.

Thanks.