Help: Functional Equation 2

Moderator's edit:

Find a continuous function $f(x)$ not everywhere zero such that $\displaystyle (f(x))^2 = \int_0^x \dfrac{f(t) \cdot \sin t}{2 + \cos t} \, dt$ .

#Calculus

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What's the problem?? Just differentiate it to get: $2 f(x) f'(x)=\dfrac{f(x) \sin x}{2+\cos x}$ Now for f(x) $\neq$ 0, we can write : $2f'(x)=\dfrac{\sin x}{2+\cos x}$ Now integrate both sides to get f(x) which is very easy...

Rishabh Jain - 5 years, 4 months ago

So it was $f^{ 2 }\left( x \right) ={ (f\left( x \right) ) }^{ 2 }$ not $f^{ 2 }\left( x \right) ={ f^{ '' }\left( x \right) }$

Akhilesh Prasad - 5 years, 4 months ago

Yeah.... Thanks to moderator's edit.... :-}

Rishabh Jain - 5 years, 4 months ago

Thanks a lot for clarifying that out had been banging my head like crazy.

Akhilesh Prasad - 5 years, 4 months ago

@parv mor

Akhilesh Prasad - 5 years, 4 months ago

@Aareyan Manzoor

Akhilesh Prasad - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$