Mixing
If $[F(x)]^{100} = int_{0}^{x} (F(t))^{100} frac{dt}{1+sin t}$ then find $F(x)$
$begingroup$
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
$$F(x) = exp(-1/(50+50tan (x/2))$$
But, I am not getting my answer right. Where did I go wrong?
calculus
edited Jan 1 at 13:15
Robert Z
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
$endgroup$
add a comment |
$begingroup$
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
$$F(x) = exp(-1/(50+50tan (x/2))$$
But, I am not getting my answer right. Where did I go wrong?
calculus
edited Jan 1 at 13:15
Robert Z
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$F(x) = exp(-1/(50+50tan (x/2))$$
But, I am not getting my answer right. Where did I go wrong?
calculus
edited Jan 1 at 13:15
Robert Z
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
$endgroup$
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
$$F(x) = exp(-1/(50+50tan (x/2))$$
But, I am not getting my answer right. Where did I go wrong?
calculus
calculus
edited Jan 1 at 13:15
Robert Z
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
edited Jan 1 at 13:15
Robert Z
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
edited Jan 1 at 13:15
Robert Z
94.2k1061132
edited Jan 1 at 13:15
Robert Z
94.2k1061132
edited Jan 1 at 13:15
Robert Z
94.2k1061132
94.2k1061132
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
asked Jan 1 at 12:49
MathsaddictMathsaddict
2908
2908
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
I think just the multiplicative constant is ommited.
$endgroup$
– user376343
Jan 1 at 13:18
$begingroup$
I think just the multiplicative constant is ommited.
$endgroup$
– user376343
Jan 1 at 13:18
1
1
$begingroup$
Actually, there is printing mistake in my answer key.
$endgroup$
– Mathsaddict
Jan 1 at 13:26
$begingroup$
Actually, there is printing mistake in my answer key.
$endgroup$
– Mathsaddict
Jan 1 at 13:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $xin (-pi,pi)$,
$$F(x)=Cexpleft(-frac{1}{50(1+tan (x/2))}right).$$
Moreover by assumption, it seems that $F(0)=0$.
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $xin (-pi,pi)$,
$$F(x)=Cexpleft(-frac{1}{50(1+tan (x/2))}right).$$
Moreover by assumption, it seems that $F(0)=0$.
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xin (-pi,pi)$,
$$F(x)=Cexpleft(-frac{1}{50(1+tan (x/2))}right).$$
Moreover by assumption, it seems that $F(0)=0$.
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xin (-pi,pi)$,
$$F(x)=Cexpleft(-frac{1}{50(1+tan (x/2))}right).$$
Moreover by assumption, it seems that $F(0)=0$.
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
$endgroup$
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 $xin (-pi,pi)$,
$$F(x)=Cexpleft(-frac{1}{50(1+tan (x/2))}right).$$
Moreover by assumption, it seems that $F(0)=0$.
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
edited Jan 1 at 13:11
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
answered Jan 1 at 12:51
Robert ZRobert Z
94.2k1061132
94.2k1061132
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
Also, $F(x)equiv 0$ should work too.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:54
$begingroup$
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
$begingroup$
@InequalitiesEverywhere Yes, you are right that is the stationary solution.
$endgroup$
– Robert Z
Jan 1 at 13:00
add a comment |
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$begingroup$
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