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?

  • $\begingroup$ 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

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 $F(0)=0$.

  • $\begingroup$ 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

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