5
$\begingroup$

If $[F(x)]^{100} = \int_{0}^{x} (F(t))^{100} \frac{dt}{1+\sin t}$ then find $F(x)$.

My attempt

Differentiating both sides,

$$100[F(x)]^{99} \frac{d F(x)}{dx} = \frac{F(x)^{100}}{1 + \sin x}$$ then $$\frac{d F(x)}{F(x)} = \frac{dx}{100(1+\sin x)}$$ and $$\int \frac{d F(x)}{F(x)} = \int \frac{dx}{100(1+\sin x)}$$ $$\log F(x) = -1/(50+50 \tan (x/2))$$ Hence $$F(x) = \exp(-1/(50+50\tan (x/2))$$ But, I am not getting my answer right. Where did I go wrong?

$\endgroup$
  • $\begingroup$ I think just the multiplicative constant is ommited. $\endgroup$ – user376343 Jan 1 at 13:18
  • 1
    $\begingroup$ Actually, there is printing mistake in my answer key. $\endgroup$ – Mathsaddict Jan 1 at 13:26
2
$\begingroup$

Note that we have the stationary solution $F(x)\equiv 0$. If $F(x)\not=0$ then, by separation of variables, $$\int \frac{d F(x)}{F(x)} = \int \frac{dx}{100(1+\sin x)} $$ which implies $$\log |F(x)| = -\frac{1}{50(1+\tan (x/2))}+c.$$ and for $x\in (-\pi,\pi)$, $$F(x)=C\exp\left(-\frac{1}{50(1+\tan (x/2))}\right).$$ Moreover by assumption, it seems that $F(0)=0$.

$\endgroup$
  • $\begingroup$ Also, $F(x)\equiv 0$ should work too. $\endgroup$ – InequalitiesEverywhere Jan 1 at 12:54
  • $\begingroup$ @InequalitiesEverywhere Yes, you are right that is the stationary solution. $\endgroup$ – Robert Z Jan 1 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.