Expert: Abe Mantell - 1/3/2004

Question
How would I go about solving this problem:

The problem gives me the solution of
y= The integral from 0 to x sin(tsquared)dt+xcubed+x+2

It also tells me that the differential equation is y(double prime)=2xcos(xsquared)+6x
and it tells me the inital conditions yprime(0)=1 and y(0)=2

How do I confirm this solution?
Thank You


Answer
Hello Molly,

Well, just take the first and second derivative of the
given solution, substitute them in to the DE and make
sure it works.

But how do you take the derivative of an integral you ask?

d/dx[Int(f(t),from t=a to x] = f(x)

So, the derivative of y is:
y' = sin(x^2) + 3x^2, now you can find y'' and check...

OK?

Abe