Daniel Ward	Posted on Friday, 25 April, 2003 - 09:44 am: They're in P6 and they're hard. Show that $y=1+\int e^{t^2}\hspace{0.5em}\mathrm{dt}$ is a solution to the differential equation ${\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\int e^{x^2}\hspace{0.5em}\mathrm{dx}}$ where $y=1$ at $x=0$. I tried differentiating the top term, but how do you differentiate the integral with respect to x ?
David Loeffler	Posted on Friday, 25 April, 2003 - 10:21 am: As for the first question, the question is what are the limits on the integral? You can't have an integral with no limits at all, it's meaningless. If it is intended to be read as $y(x)=1+{\int }_0^x{e}^{t^2}\hspace{0.5em}\mathrm{dt}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=e^{x^2}$ obviously (by the fundamental theorem of calculus). Note the absence of an integral sign here. David
Daniel Ward	Posted on Friday, 25 April, 2003 - 10:57 pm: Yup, sorry for the extra integral sign and the misplaced limits. One day I'll learn how to proofread properly :-) I'm afraid I don't understand your explanation. Could you explain a little more ? Dan
David Loeffler	Posted on Saturday, 26 April, 2003 - 08:18 am: The fundamental theorem of calculus, which I invoked for the first question, is just the statement that differentiation and integration are inverse operations; if F(x) = Ã????² x0 x f(t) dt, then F'(x) = f(x). David