$f'(\sin^2x) = \cos^2x + \tan^2x,\quad$ subject to $\quad0 Many thanks.
How do I go about finding some $f(x)$ that satisfies the following first-order DE?
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calculus
ordinary-differential-equations
trigonometry
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0I have seen this question before not so long ago on this site ..so voting to close. – 2017-02-11
2 Answers
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Note that $1 = \cos^2 x + \sin^2 x$ and $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ So for $y = \sin^2 x$ we get:
$$f'(y) = (1-y) + \frac{1}{1-y}$$
for $y \in \text{Range}(\sin^2 x) = (0,1)$ this equation can be solved using straight forward integration (i.e. you're looking for an antiderivative of $f'(y)$.)
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$f'(\sin^2x) = \cos^2x + \tan^2x= (1-sin^2x)+(\frac{sin^2x}{1-sin^2x})$
$\implies f(sin^2x)=(sin^2x-\frac{1}{2}sin^4x)+(-sin^2x-ln(1-sin^2x))+C$
I hope I'm correct