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Is it possible find a function $u(x)$ so that $[y'(x)+y(x)\tan(x)]^2=(u'(x))^2-(u(x))^2$?

If not, is there an obvious reason why the integrals of the LHS an the RHS respectively over the interval $(0,1)$ are equal?

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Is this of any help? $$u(x) = y(x) \sec(x)\\ u'(x) = y'(x) \sec(x) + y(x) \sec(x) \tan(x)$$ $$\left(u'(x) \right)^2 - (u(x))^2= (y'(x) + y(x) \tan(x))^2 \sec^2(x) - y^2(x) \sec^2(x)$$ –  user17762 May 23 '12 at 21:05
@Marvis: Thanks for the suggestion! –  freda johnson May 23 '12 at 21:27
but at the moment i am still not entirely sure of how this works. maybe integrating by parts would help? –  freda johnson May 23 '12 at 21:31
I don't understand your second question. The integrals of what over that interval are equal? –  Greg Martin May 23 '12 at 22:17
@GregMartin: the LHS and the RHS respectively. –  freda johnson May 23 '12 at 22:26

1 Answer 1

up vote 1 down vote accepted

To get a suitable $u$, you might solve the differential equation $u'(x) = \sqrt{(y'(x)+y(x) \tan(x))^2 + u(x)^2}$. Assuming $y$ and $y'$ are continuous on $[0,1]$, the right side of that differential equation is continuous and locally Lipschitz on $[0,1] \times \mathbb R$, so for any initial condition at some $x_0 \in [0,1]$ we have local existence and uniqueness of solutions. Moreover, since $\sqrt{A^2 + u^2} \le |A| + |u|$, the solution will exist on all of $[0,1]$.

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Thank you very much! –  freda johnson May 24 '12 at 5:36

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