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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? $$u(x) = y(x) \sec(x)\\ u'(x) = y'(x) \sec(x) + y(x) \sec(x) \tan(x)$$ $$\left(u'(x) \right)^2 - (u(x))^2= (y'(x) + y(x) \tan(x))^2 \sec^2(x) - y^2(x) \sec^2(x)$$2012-05-23
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    @Marvis: Thanks for the suggestion!2012-05-23
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    but at the moment i am still not entirely sure of how this works. maybe integrating by parts would help?2012-05-23
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    I don't understand your second question. The integrals of *what* over that interval are equal?2012-05-23
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    @GregMartin: the LHS and the RHS respectively.2012-05-23
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    @GregMartin Left hand side of the mentioned equality and right hand side of it!2012-05-24
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    If the left and right sides of the equation are equal everywhere on $[0,1]$ and are integrable over $[0,1]$, their integrals are equal. If the left and right sides are not equal, what is supposed to be the relation between $y$ and $u$?2012-05-24

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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 $[0,1]$.