Differential equation involving inverse function theorem

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Pardon me if you find the question crazy.

How do we solve the differential equation,

$$fracdf^-1(t)dt Biggm |_f(t) = frac1f(t)(1-f(t))$$.

without using the following approach:

If we multiply LHS and RHS by $df$, by inverse function theorem we have for LHS,
$$ fracdf^-1(t)dt Biggm |_f(t) df = dt $$

And thus,

$$ fracdfdt = f(t)(1-f(t))$$ and we can solve this.

differential-equations

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asked Jul 19 at 2:18

user2167741

1039

add a comment | 

up vote
3
down vote

favorite

Pardon me if you find the question crazy.

How do we solve the differential equation,

$$fracdf^-1(t)dt Biggm |_f(t) = frac1f(t)(1-f(t))$$.

without using the following approach:

If we multiply LHS and RHS by $df$, by inverse function theorem we have for LHS,
$$ fracdf^-1(t)dt Biggm |_f(t) df = dt $$

And thus,

$$ fracdfdt = f(t)(1-f(t))$$ and we can solve this.

differential-equations

share|cite|improve this question

asked Jul 19 at 2:18

user2167741

1039

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Pardon me if you find the question crazy.

How do we solve the differential equation,

$$fracdf^-1(t)dt Biggm |_f(t) = frac1f(t)(1-f(t))$$.

without using the following approach:

If we multiply LHS and RHS by $df$, by inverse function theorem we have for LHS,
$$ fracdf^-1(t)dt Biggm |_f(t) df = dt $$

And thus,

$$ fracdfdt = f(t)(1-f(t))$$ and we can solve this.

differential-equations

share|cite|improve this question

asked Jul 19 at 2:18

user2167741

1039

Pardon me if you find the question crazy.

How do we solve the differential equation,

$$fracdf^-1(t)dt Biggm |_f(t) = frac1f(t)(1-f(t))$$.

without using the following approach:

If we multiply LHS and RHS by $df$, by inverse function theorem we have for LHS,
$$ fracdf^-1(t)dt Biggm |_f(t) df = dt $$

And thus,

$$ fracdfdt = f(t)(1-f(t))$$ and we can solve this.

differential-equations

share|cite|improve this question

asked Jul 19 at 2:18

user2167741

1039

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 2:18

user2167741

1039

asked Jul 19 at 2:18

user2167741

1039

asked Jul 19 at 2:18

user2167741

1039

1039

add a comment | 

add a comment | 

1 Answer
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Notice that the notation $$fracdf^-1(t)dtlvert_f(t)$$ means $$fracdf^-1(f(t))df(t)=fracdtdf(t)$$

So we obtain
$$fracdtdf=frac1fcdot(1-f)$$
which is separable: $$dt=fracdffcdot(1-f)implies t+C=ln|fcdot(1-f)|$$

share|cite|improve this answer

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
  – user2167741
  Jul 19 at 2:37
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Notice that the notation $$fracdf^-1(t)dtlvert_f(t)$$ means $$fracdf^-1(f(t))df(t)=fracdtdf(t)$$

So we obtain
$$fracdtdf=frac1fcdot(1-f)$$
which is separable: $$dt=fracdffcdot(1-f)implies t+C=ln|fcdot(1-f)|$$

share|cite|improve this answer

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
  – user2167741
  Jul 19 at 2:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

Notice that the notation $$fracdf^-1(t)dtlvert_f(t)$$ means $$fracdf^-1(f(t))df(t)=fracdtdf(t)$$

So we obtain
$$fracdtdf=frac1fcdot(1-f)$$
which is separable: $$dt=fracdffcdot(1-f)implies t+C=ln|fcdot(1-f)|$$

share|cite|improve this answer

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
  – user2167741
  Jul 19 at 2:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Notice that the notation $$fracdf^-1(t)dtlvert_f(t)$$ means $$fracdf^-1(f(t))df(t)=fracdtdf(t)$$

So we obtain
$$fracdtdf=frac1fcdot(1-f)$$
which is separable: $$dt=fracdffcdot(1-f)implies t+C=ln|fcdot(1-f)|$$

share|cite|improve this answer

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

Notice that the notation $$fracdf^-1(t)dtlvert_f(t)$$ means $$fracdf^-1(f(t))df(t)=fracdtdf(t)$$

So we obtain
$$fracdtdf=frac1fcdot(1-f)$$
which is separable: $$dt=fracdffcdot(1-f)implies t+C=ln|fcdot(1-f)|$$

share|cite|improve this answer

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 7:19

edited Jul 19 at 7:19

edited Jul 19 at 7:19

answered Jul 19 at 2:30

Szeto

4,1981521

answered Jul 19 at 2:30

Szeto

4,1981521

answered Jul 19 at 2:30

Szeto

4,1981521

4,1981521

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
  – user2167741
  Jul 19 at 2:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
  – user2167741
  Jul 19 at 2:37
  

  
  
  
  

Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
– user2167741
Jul 19 at 2:37

Well isn't the notation $$fracdf^-1(t)dt Biggm |_f(t)$$ mean first taking the derivative with respect to $t$ and substituting $f(t)$ than just substituting like what you did?
– user2167741
Jul 19 at 2:37

add a comment | 

 

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