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

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So this is what I did. I integrated the function to get,
Then took the derivative of that with respect to x to get

Then I sub it in to get which does not equal 0, so i'm saying that this function is not a solution to the differential equation.

IS this correct?
David.

I see no reason to integrate when you just turn around and differentiate. With by the "Fundamental Theorem of Calculus", that the derivative of is just f(x), we get immediately your . And that does NOT satisfy the differential equation, again as you say.

HOWEVER, I think you have misread the problem. If , 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 . 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.)

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.