how to solve this first order nonlinear differential equation

2

$begingroup$

I'm reading nonlinear control systems book. The author provides this example
$$
dot{x} = r + x^2, quad r < 0.
$$
I would like to compute the analytical solution for the proceeding ODE. My attempt is

$$
begin{align}
frac{dx}{dt} &= r + x^2 \
frac{dx}{r+x^2} &= dt \
int^{x(t)}_{x_0} frac{1}{r+x^2} dx &= int^{t}_{t_0} dtau \
frac{tan^{-1}left(frac{x}{sqrt{r}}right)}{sqrt{r}} Big|^{x(t)}_{x_0} &= (t-t_0)
end{align}
$$

Now the problem with the assumption that $r<0$, how I can handle the substitution for the left side? I need to reach the final step where $x(t)$ is solely in the left side.

ordinary-differential-equations nonlinear-system

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asked Jan 30 at 6:07

CroCoCroCo

271221

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
  $endgroup$
  – Christoph
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
  $endgroup$
  – Arctic Char
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
  $endgroup$
  – CroCo
  Jan 30 at 6:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $r<0$ you have $-r>0$, so you can take the square root of that.
  $endgroup$
  – Christoph
  Jan 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $-a^2=r$, so $a$ is the square root of the positive number $-r$.
  $endgroup$
  – Hans Lundmark
  Jan 30 at 6:23
  

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

I'm reading nonlinear control systems book. The author provides this example
$$
dot{x} = r + x^2, quad r < 0.
$$
I would like to compute the analytical solution for the proceeding ODE. My attempt is

$$
begin{align}
frac{dx}{dt} &= r + x^2 \
frac{dx}{r+x^2} &= dt \
int^{x(t)}_{x_0} frac{1}{r+x^2} dx &= int^{t}_{t_0} dtau \
frac{tan^{-1}left(frac{x}{sqrt{r}}right)}{sqrt{r}} Big|^{x(t)}_{x_0} &= (t-t_0)
end{align}
$$

Now the problem with the assumption that $r<0$, how I can handle the substitution for the left side? I need to reach the final step where $x(t)$ is solely in the left side.

ordinary-differential-equations nonlinear-system

share|cite|improve this question

asked Jan 30 at 6:07

CroCoCroCo

271221

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
  $endgroup$
  – Christoph
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
  $endgroup$
  – Arctic Char
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
  $endgroup$
  – CroCo
  Jan 30 at 6:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $r<0$ you have $-r>0$, so you can take the square root of that.
  $endgroup$
  – Christoph
  Jan 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $-a^2=r$, so $a$ is the square root of the positive number $-r$.
  $endgroup$
  – Hans Lundmark
  Jan 30 at 6:23
  

  
  
  
  

 | 
show 3 more comments

2

2

2

$begingroup$

I'm reading nonlinear control systems book. The author provides this example
$$
dot{x} = r + x^2, quad r < 0.
$$
I would like to compute the analytical solution for the proceeding ODE. My attempt is

$$
begin{align}
frac{dx}{dt} &= r + x^2 \
frac{dx}{r+x^2} &= dt \
int^{x(t)}_{x_0} frac{1}{r+x^2} dx &= int^{t}_{t_0} dtau \
frac{tan^{-1}left(frac{x}{sqrt{r}}right)}{sqrt{r}} Big|^{x(t)}_{x_0} &= (t-t_0)
end{align}
$$

Now the problem with the assumption that $r<0$, how I can handle the substitution for the left side? I need to reach the final step where $x(t)$ is solely in the left side.

ordinary-differential-equations nonlinear-system

share|cite|improve this question

asked Jan 30 at 6:07

CroCoCroCo

271221

$endgroup$

I'm reading nonlinear control systems book. The author provides this example
$$
dot{x} = r + x^2, quad r < 0.
$$
I would like to compute the analytical solution for the proceeding ODE. My attempt is

$$
begin{align}
frac{dx}{dt} &= r + x^2 \
frac{dx}{r+x^2} &= dt \
int^{x(t)}_{x_0} frac{1}{r+x^2} dx &= int^{t}_{t_0} dtau \
frac{tan^{-1}left(frac{x}{sqrt{r}}right)}{sqrt{r}} Big|^{x(t)}_{x_0} &= (t-t_0)
end{align}
$$

Now the problem with the assumption that $r<0$, how I can handle the substitution for the left side? I need to reach the final step where $x(t)$ is solely in the left side.

ordinary-differential-equations nonlinear-system

ordinary-differential-equations nonlinear-system

share|cite|improve this question

asked Jan 30 at 6:07

CroCoCroCo

271221

share|cite|improve this question

asked Jan 30 at 6:07

CroCoCroCo

271221

share|cite|improve this question

share|cite|improve this question

asked Jan 30 at 6:07

CroCoCroCo

271221

asked Jan 30 at 6:07

CroCoCroCo

271221

asked Jan 30 at 6:07

CroCoCroCo

271221

271221

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
  $endgroup$
  – Christoph
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
  $endgroup$
  – Arctic Char
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
  $endgroup$
  – CroCo
  Jan 30 at 6:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $r<0$ you have $-r>0$, so you can take the square root of that.
  $endgroup$
  – Christoph
  Jan 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $-a^2=r$, so $a$ is the square root of the positive number $-r$.
  $endgroup$
  – Hans Lundmark
  Jan 30 at 6:23
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
  $endgroup$
  – Christoph
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
  $endgroup$
  – Arctic Char
  Jan 30 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
  $endgroup$
  – CroCo
  Jan 30 at 6:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $r<0$ you have $-r>0$, so you can take the square root of that.
  $endgroup$
  – Christoph
  Jan 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $-a^2=r$, so $a$ is the square root of the positive number $-r$.
  $endgroup$
  – Hans Lundmark
  Jan 30 at 6:23
  

  
  
  
  

1

1

$begingroup$
Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
$endgroup$
– Christoph
Jan 30 at 6:16

$begingroup$
Substitute $u = frac{x}{sqrt{-r}}$. You'll get an inverse hyperbolic tangent.
$endgroup$
– Christoph
Jan 30 at 6:16

1

1

$begingroup$
So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
$endgroup$
– Arctic Char
Jan 30 at 6:16

$begingroup$
So it is $x^2 - a^2$ at the bottom. Could use partial fractions.
$endgroup$
– Arctic Char
Jan 30 at 6:16

$begingroup$
@Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
$endgroup$
– CroCo
Jan 30 at 6:21

$begingroup$
@Christoph, why did you put the minus ? Also, how to handle the square root of negative number?
$endgroup$
– CroCo
Jan 30 at 6:21

1

1

$begingroup$
If $r<0$ you have $-r>0$, so you can take the square root of that.
$endgroup$
– Christoph
Jan 30 at 6:22

$begingroup$
If $r<0$ you have $-r>0$, so you can take the square root of that.
$endgroup$
– Christoph
Jan 30 at 6:22

1

1

$begingroup$
$-a^2=r$, so $a$ is the square root of the positive number $-r$.
$endgroup$
– Hans Lundmark
Jan 30 at 6:23

$begingroup$
$-a^2=r$, so $a$ is the square root of the positive number $-r$.
$endgroup$
– Hans Lundmark
Jan 30 at 6:23

 | 
show 3 more comments

4 Answers
4

active

oldest

votes

2

$begingroup$

Let $r=-a^2$. Then
$$intfrac{dx}{x^2-a^2}=int dt$$
$$-frac{log{left( x+aright) }-log{left( x-aright) }}{2 a}=t+c$$
Take $c=frac{log C}{2a}$.
$$log{left( frac{x-a}{C, left( x+aright) }right) }=2 a t,$$
$$frac{x-a}{x+a}=C, {{e}^{2 a t}}.$$
General solution is
$$x=frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

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add a comment | 

5

$begingroup$

Let us consider

$$dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$frac{dx}{1-x^2}=-dt$$ and

$$text{artanh }x-text{artanh }x_0=t_0-t,$$

i.e.

$$x=tanh(t_0-t+text{artanh }x_0).$$

When $|x|>1$, we solve with

$$text{arcoth }x-text{arcoth }x_0=t_0-t,$$

i.e.

$$x=coth(t_0-t+text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+text{arcoth }x_0$.

Finally, $x=pm1$ are two valid solutions.

share|cite|improve this answer

edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

$endgroup$

add a comment | 

3

$begingroup$

Solve it as Riccati equation by setting $x=-frac{u'}{u}$. Then
$$
u''-a^2u=0
$$
has the solution $u(t)=Ce^{at}+De^{-at}$ and thus
$$
y(x)=-afrac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}}
$$
with some redundancy in the parameter pair $(C,D)$.

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answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

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add a comment | 

0

$begingroup$

We use separation of variables to solve the ODE $dot{x} = r + x^2$, $r < 0$:

1. We first find the constant solutions $x equiv pm sqrt{-r}$. In the following steps we shall assume $r+x^2 not equiv 0$.




2. We use the Leibniz notation and we separate the variables: $frac{dx}{dt} = r + x^2$, $frac{1}{r+x^2} dx = dt$.




3. We integrate on both sides: $int frac{1}{r+x^2} , mathrm{d}x = int 1 , mathrm{d}t$.




4. We solve the integral on the left-hand side using the substitution $x = sqrt{-r} u$, $dx = sqrt{-r} du$:
   begin{eqnarray}
   int frac{1}{r+x^2} , mathrm{d}x &=& int frac{1}{r-ru^2} sqrt{-r} , mathrm{d}u = frac{sqrt{-r}}{r} int frac{1}{1-u^2} , mathrm{d}u = frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}(u) + tilde{C}, & |u| < 1\
   operatorname{arcoth}(u) + tilde{C}, & |u| > 1
   end{array}
   right.\
   &=& frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| < sqrt{-r}\
   operatorname{arcoth}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| > sqrt{-r}
   end{array}
   right., quad tilde{C} in mathbb{R}.
   end{eqnarray}

Thus we now obtain the nonlinear equations
begin{eqnarray}
operatorname{artanh}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| < sqrt{-r},\
operatorname{arcoth}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| > sqrt{-r},
end{eqnarray}
with a constant $C in mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 in mathbb{R}$:
begin{eqnarray}
x(t) &=& sqrt{-r} cothleft(operatorname{arcoth}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0)right), quad |x_0| > sqrt{-r},\
x(t) &=& -sqrt{-r}, quad x_0 = -sqrt{-r},\
x(t) &=& sqrt{-r}, quad x_0 = sqrt{-r},\
x(t) &=& sqrt{-r} tanhleft(operatorname{artanh}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0) right), quad |x_0| < sqrt{-r}.
end{eqnarray}

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answered Jan 30 at 9:39

ChristophChristoph

59616

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add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Let $r=-a^2$. Then
$$intfrac{dx}{x^2-a^2}=int dt$$
$$-frac{log{left( x+aright) }-log{left( x-aright) }}{2 a}=t+c$$
Take $c=frac{log C}{2a}$.
$$log{left( frac{x-a}{C, left( x+aright) }right) }=2 a t,$$
$$frac{x-a}{x+a}=C, {{e}^{2 a t}}.$$
General solution is
$$x=frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

add a comment | 

2

$begingroup$

Let $r=-a^2$. Then
$$intfrac{dx}{x^2-a^2}=int dt$$
$$-frac{log{left( x+aright) }-log{left( x-aright) }}{2 a}=t+c$$
Take $c=frac{log C}{2a}$.
$$log{left( frac{x-a}{C, left( x+aright) }right) }=2 a t,$$
$$frac{x-a}{x+a}=C, {{e}^{2 a t}}.$$
General solution is
$$x=frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

add a comment | 

2

2

2

$begingroup$

Let $r=-a^2$. Then
$$intfrac{dx}{x^2-a^2}=int dt$$
$$-frac{log{left( x+aright) }-log{left( x-aright) }}{2 a}=t+c$$
Take $c=frac{log C}{2a}$.
$$log{left( frac{x-a}{C, left( x+aright) }right) }=2 a t,$$
$$frac{x-a}{x+a}=C, {{e}^{2 a t}}.$$
General solution is
$$x=frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

Let $r=-a^2$. Then
$$intfrac{dx}{x^2-a^2}=int dt$$
$$-frac{log{left( x+aright) }-log{left( x-aright) }}{2 a}=t+c$$
Take $c=frac{log C}{2a}$.
$$log{left( frac{x-a}{C, left( x+aright) }right) }=2 a t,$$
$$frac{x-a}{x+a}=C, {{e}^{2 a t}}.$$
General solution is
$$x=frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

answered Jan 30 at 6:41

Aleksas DomarkasAleksas Domarkas

1,62317

1,62317

add a comment | 

add a comment | 

5

$begingroup$

Let us consider

$$dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$frac{dx}{1-x^2}=-dt$$ and

$$text{artanh }x-text{artanh }x_0=t_0-t,$$

i.e.

$$x=tanh(t_0-t+text{artanh }x_0).$$

When $|x|>1$, we solve with

$$text{arcoth }x-text{arcoth }x_0=t_0-t,$$

i.e.

$$x=coth(t_0-t+text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+text{arcoth }x_0$.

Finally, $x=pm1$ are two valid solutions.

share|cite|improve this answer

edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

$endgroup$

add a comment | 

5

$begingroup$

Let us consider

$$dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$frac{dx}{1-x^2}=-dt$$ and

$$text{artanh }x-text{artanh }x_0=t_0-t,$$

i.e.

$$x=tanh(t_0-t+text{artanh }x_0).$$

When $|x|>1$, we solve with

$$text{arcoth }x-text{arcoth }x_0=t_0-t,$$

i.e.

$$x=coth(t_0-t+text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+text{arcoth }x_0$.

Finally, $x=pm1$ are two valid solutions.

share|cite|improve this answer

edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

$endgroup$

add a comment | 

5

5

5

$begingroup$

Let us consider

$$dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$frac{dx}{1-x^2}=-dt$$ and

$$text{artanh }x-text{artanh }x_0=t_0-t,$$

i.e.

$$x=tanh(t_0-t+text{artanh }x_0).$$

When $|x|>1$, we solve with

$$text{arcoth }x-text{arcoth }x_0=t_0-t,$$

i.e.

$$x=coth(t_0-t+text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+text{arcoth }x_0$.

Finally, $x=pm1$ are two valid solutions.

share|cite|improve this answer

edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

$endgroup$

Let us consider

$$dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$frac{dx}{1-x^2}=-dt$$ and

$$text{artanh }x-text{artanh }x_0=t_0-t,$$

i.e.

$$x=tanh(t_0-t+text{artanh }x_0).$$

When $|x|>1$, we solve with

$$text{arcoth }x-text{arcoth }x_0=t_0-t,$$

i.e.

$$x=coth(t_0-t+text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+text{arcoth }x_0$.

Finally, $x=pm1$ are two valid solutions.

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edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

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share|cite|improve this answer

edited Jan 30 at 10:14

edited Jan 30 at 10:14

edited Jan 30 at 10:14

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

answered Jan 30 at 10:03

Yves DaoustYves Daoust

132k676229

132k676229

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add a comment | 

3

$begingroup$

Solve it as Riccati equation by setting $x=-frac{u'}{u}$. Then
$$
u''-a^2u=0
$$
has the solution $u(t)=Ce^{at}+De^{-at}$ and thus
$$
y(x)=-afrac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}}
$$
with some redundancy in the parameter pair $(C,D)$.

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answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

$endgroup$

add a comment | 

3

$begingroup$

Solve it as Riccati equation by setting $x=-frac{u'}{u}$. Then
$$
u''-a^2u=0
$$
has the solution $u(t)=Ce^{at}+De^{-at}$ and thus
$$
y(x)=-afrac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}}
$$
with some redundancy in the parameter pair $(C,D)$.

share|cite|improve this answer

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

$endgroup$

add a comment | 

3

3

3

$begingroup$

Solve it as Riccati equation by setting $x=-frac{u'}{u}$. Then
$$
u''-a^2u=0
$$
has the solution $u(t)=Ce^{at}+De^{-at}$ and thus
$$
y(x)=-afrac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}}
$$
with some redundancy in the parameter pair $(C,D)$.

share|cite|improve this answer

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

$endgroup$

Solve it as Riccati equation by setting $x=-frac{u'}{u}$. Then
$$
u''-a^2u=0
$$
has the solution $u(t)=Ce^{at}+De^{-at}$ and thus
$$
y(x)=-afrac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}}
$$
with some redundancy in the parameter pair $(C,D)$.

share|cite|improve this answer

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

answered Jan 30 at 8:09

LutzLLutzL

60.2k42057

60.2k42057

add a comment | 

add a comment | 

0

$begingroup$

We use separation of variables to solve the ODE $dot{x} = r + x^2$, $r < 0$:

1. We first find the constant solutions $x equiv pm sqrt{-r}$. In the following steps we shall assume $r+x^2 not equiv 0$.




2. We use the Leibniz notation and we separate the variables: $frac{dx}{dt} = r + x^2$, $frac{1}{r+x^2} dx = dt$.




3. We integrate on both sides: $int frac{1}{r+x^2} , mathrm{d}x = int 1 , mathrm{d}t$.




4. We solve the integral on the left-hand side using the substitution $x = sqrt{-r} u$, $dx = sqrt{-r} du$:
   begin{eqnarray}
   int frac{1}{r+x^2} , mathrm{d}x &=& int frac{1}{r-ru^2} sqrt{-r} , mathrm{d}u = frac{sqrt{-r}}{r} int frac{1}{1-u^2} , mathrm{d}u = frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}(u) + tilde{C}, & |u| < 1\
   operatorname{arcoth}(u) + tilde{C}, & |u| > 1
   end{array}
   right.\
   &=& frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| < sqrt{-r}\
   operatorname{arcoth}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| > sqrt{-r}
   end{array}
   right., quad tilde{C} in mathbb{R}.
   end{eqnarray}

Thus we now obtain the nonlinear equations
begin{eqnarray}
operatorname{artanh}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| < sqrt{-r},\
operatorname{arcoth}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| > sqrt{-r},
end{eqnarray}
with a constant $C in mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 in mathbb{R}$:
begin{eqnarray}
x(t) &=& sqrt{-r} cothleft(operatorname{arcoth}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0)right), quad |x_0| > sqrt{-r},\
x(t) &=& -sqrt{-r}, quad x_0 = -sqrt{-r},\
x(t) &=& sqrt{-r}, quad x_0 = sqrt{-r},\
x(t) &=& sqrt{-r} tanhleft(operatorname{artanh}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0) right), quad |x_0| < sqrt{-r}.
end{eqnarray}

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answered Jan 30 at 9:39

ChristophChristoph

59616

$endgroup$

add a comment | 

0

$begingroup$

We use separation of variables to solve the ODE $dot{x} = r + x^2$, $r < 0$:

1. We first find the constant solutions $x equiv pm sqrt{-r}$. In the following steps we shall assume $r+x^2 not equiv 0$.




2. We use the Leibniz notation and we separate the variables: $frac{dx}{dt} = r + x^2$, $frac{1}{r+x^2} dx = dt$.




3. We integrate on both sides: $int frac{1}{r+x^2} , mathrm{d}x = int 1 , mathrm{d}t$.




4. We solve the integral on the left-hand side using the substitution $x = sqrt{-r} u$, $dx = sqrt{-r} du$:
   begin{eqnarray}
   int frac{1}{r+x^2} , mathrm{d}x &=& int frac{1}{r-ru^2} sqrt{-r} , mathrm{d}u = frac{sqrt{-r}}{r} int frac{1}{1-u^2} , mathrm{d}u = frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}(u) + tilde{C}, & |u| < 1\
   operatorname{arcoth}(u) + tilde{C}, & |u| > 1
   end{array}
   right.\
   &=& frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| < sqrt{-r}\
   operatorname{arcoth}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| > sqrt{-r}
   end{array}
   right., quad tilde{C} in mathbb{R}.
   end{eqnarray}

Thus we now obtain the nonlinear equations
begin{eqnarray}
operatorname{artanh}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| < sqrt{-r},\
operatorname{arcoth}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| > sqrt{-r},
end{eqnarray}
with a constant $C in mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 in mathbb{R}$:
begin{eqnarray}
x(t) &=& sqrt{-r} cothleft(operatorname{arcoth}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0)right), quad |x_0| > sqrt{-r},\
x(t) &=& -sqrt{-r}, quad x_0 = -sqrt{-r},\
x(t) &=& sqrt{-r}, quad x_0 = sqrt{-r},\
x(t) &=& sqrt{-r} tanhleft(operatorname{artanh}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0) right), quad |x_0| < sqrt{-r}.
end{eqnarray}

share|cite|improve this answer

answered Jan 30 at 9:39

ChristophChristoph

59616

$endgroup$

add a comment | 

0

0

0

$begingroup$

We use separation of variables to solve the ODE $dot{x} = r + x^2$, $r < 0$:

1. We first find the constant solutions $x equiv pm sqrt{-r}$. In the following steps we shall assume $r+x^2 not equiv 0$.




2. We use the Leibniz notation and we separate the variables: $frac{dx}{dt} = r + x^2$, $frac{1}{r+x^2} dx = dt$.




3. We integrate on both sides: $int frac{1}{r+x^2} , mathrm{d}x = int 1 , mathrm{d}t$.




4. We solve the integral on the left-hand side using the substitution $x = sqrt{-r} u$, $dx = sqrt{-r} du$:
   begin{eqnarray}
   int frac{1}{r+x^2} , mathrm{d}x &=& int frac{1}{r-ru^2} sqrt{-r} , mathrm{d}u = frac{sqrt{-r}}{r} int frac{1}{1-u^2} , mathrm{d}u = frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}(u) + tilde{C}, & |u| < 1\
   operatorname{arcoth}(u) + tilde{C}, & |u| > 1
   end{array}
   right.\
   &=& frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| < sqrt{-r}\
   operatorname{arcoth}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| > sqrt{-r}
   end{array}
   right., quad tilde{C} in mathbb{R}.
   end{eqnarray}

Thus we now obtain the nonlinear equations
begin{eqnarray}
operatorname{artanh}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| < sqrt{-r},\
operatorname{arcoth}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| > sqrt{-r},
end{eqnarray}
with a constant $C in mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 in mathbb{R}$:
begin{eqnarray}
x(t) &=& sqrt{-r} cothleft(operatorname{arcoth}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0)right), quad |x_0| > sqrt{-r},\
x(t) &=& -sqrt{-r}, quad x_0 = -sqrt{-r},\
x(t) &=& sqrt{-r}, quad x_0 = sqrt{-r},\
x(t) &=& sqrt{-r} tanhleft(operatorname{artanh}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0) right), quad |x_0| < sqrt{-r}.
end{eqnarray}

share|cite|improve this answer

answered Jan 30 at 9:39

ChristophChristoph

59616

$endgroup$

We use separation of variables to solve the ODE $dot{x} = r + x^2$, $r < 0$:

1. We first find the constant solutions $x equiv pm sqrt{-r}$. In the following steps we shall assume $r+x^2 not equiv 0$.




2. We use the Leibniz notation and we separate the variables: $frac{dx}{dt} = r + x^2$, $frac{1}{r+x^2} dx = dt$.




3. We integrate on both sides: $int frac{1}{r+x^2} , mathrm{d}x = int 1 , mathrm{d}t$.




4. We solve the integral on the left-hand side using the substitution $x = sqrt{-r} u$, $dx = sqrt{-r} du$:
   begin{eqnarray}
   int frac{1}{r+x^2} , mathrm{d}x &=& int frac{1}{r-ru^2} sqrt{-r} , mathrm{d}u = frac{sqrt{-r}}{r} int frac{1}{1-u^2} , mathrm{d}u = frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}(u) + tilde{C}, & |u| < 1\
   operatorname{arcoth}(u) + tilde{C}, & |u| > 1
   end{array}
   right.\
   &=& frac{sqrt{-r}}{r} left{ begin{array}{ll}
   operatorname{artanh}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| < sqrt{-r}\
   operatorname{arcoth}left(frac{x}{sqrt{-r}}right) + tilde{C}, & |x| > sqrt{-r}
   end{array}
   right., quad tilde{C} in mathbb{R}.
   end{eqnarray}

Thus we now obtain the nonlinear equations
begin{eqnarray}
operatorname{artanh}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| < sqrt{-r},\
operatorname{arcoth}left(frac{x}{sqrt{-r}}right) = C -sqrt{-r} t, quad |x| > sqrt{-r},
end{eqnarray}
with a constant $C in mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 in mathbb{R}$:
begin{eqnarray}
x(t) &=& sqrt{-r} cothleft(operatorname{arcoth}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0)right), quad |x_0| > sqrt{-r},\
x(t) &=& -sqrt{-r}, quad x_0 = -sqrt{-r},\
x(t) &=& sqrt{-r}, quad x_0 = sqrt{-r},\
x(t) &=& sqrt{-r} tanhleft(operatorname{artanh}left(frac{x_0}{sqrt{-r}}right)-sqrt{-r}(t-t_0) right), quad |x_0| < sqrt{-r}.
end{eqnarray}

share|cite|improve this answer

answered Jan 30 at 9:39

ChristophChristoph

59616

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 9:39

ChristophChristoph

59616

answered Jan 30 at 9:39

ChristophChristoph

59616

answered Jan 30 at 9:39

ChristophChristoph

59616

59616

add a comment | 

add a comment | 

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