How to solve this kind of nonlinear differential equation?

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up vote
3
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I have to solve the following equation

$$ z'-e^z=2$$

where $z=f(x)$, but I can't solve this. Could you please help me ?

differential-equations

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edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

asked Aug 19 at 11:46

MysteryGuy

377216

add a comment | 

up vote
3
down vote

favorite

I have to solve the following equation

$$ z'-e^z=2$$

where $z=f(x)$, but I can't solve this. Could you please help me ?

differential-equations

share|cite|improve this question

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

asked Aug 19 at 11:46

MysteryGuy

377216

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I have to solve the following equation

$$ z'-e^z=2$$

where $z=f(x)$, but I can't solve this. Could you please help me ?

differential-equations

share|cite|improve this question

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

asked Aug 19 at 11:46

MysteryGuy

377216

I have to solve the following equation

$$ z'-e^z=2$$

where $z=f(x)$, but I can't solve this. Could you please help me ?

differential-equations

share|cite|improve this question

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

asked Aug 19 at 11:46

MysteryGuy

377216

share|cite|improve this question

share|cite|improve this question

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

edited Aug 19 at 11:52

Rodrigo de Azevedo

12.7k41751

12.7k41751

asked Aug 19 at 11:46

MysteryGuy

377216

asked Aug 19 at 11:46

MysteryGuy

377216

asked Aug 19 at 11:46

MysteryGuy

377216

377216

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
8
down vote



accepted

Try multiplying by $mathrme^-z$, as this will lead to
$$
mathrme^-zz’ - 1 = 2mathrme^-z
$$
Or
$$
-(mathrme^-z)’ - 1 = 2mathrme^-z
$$
We can then re-write as
$$
-y’ -1 = 2y
$$
With $y=mathrme^-z$. Should be easier to solve.

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

add a comment | 

up vote
1
down vote

it looks like Bernouilli 's equation
$$z'=e^z+2$$
substitute
$$u=e^z+2 implies frac dudz=e^z=u-2$$
The equation becomes
$$frac dzdt=u$$
$$frac dzdufrac dudt=u$$
$$frac dudt=ufrac dudz$$
$$frac dudt=u(u-2)$$
$$u'_t+2u=u^2$$
This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way
$$int frac duu(u-2)=int dt$$

share|cite|improve this answer

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
8
down vote



accepted

Try multiplying by $mathrme^-z$, as this will lead to
$$
mathrme^-zz’ - 1 = 2mathrme^-z
$$
Or
$$
-(mathrme^-z)’ - 1 = 2mathrme^-z
$$
We can then re-write as
$$
-y’ -1 = 2y
$$
With $y=mathrme^-z$. Should be easier to solve.

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

add a comment | 

up vote
8
down vote



accepted

Try multiplying by $mathrme^-z$, as this will lead to
$$
mathrme^-zz’ - 1 = 2mathrme^-z
$$
Or
$$
-(mathrme^-z)’ - 1 = 2mathrme^-z
$$
We can then re-write as
$$
-y’ -1 = 2y
$$
With $y=mathrme^-z$. Should be easier to solve.

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

add a comment | 

up vote
8
down vote



accepted

up vote
8
down vote



accepted

Try multiplying by $mathrme^-z$, as this will lead to
$$
mathrme^-zz’ - 1 = 2mathrme^-z
$$
Or
$$
-(mathrme^-z)’ - 1 = 2mathrme^-z
$$
We can then re-write as
$$
-y’ -1 = 2y
$$
With $y=mathrme^-z$. Should be easier to solve.

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

Try multiplying by $mathrme^-z$, as this will lead to
$$
mathrme^-zz’ - 1 = 2mathrme^-z
$$
Or
$$
-(mathrme^-z)’ - 1 = 2mathrme^-z
$$
We can then re-write as
$$
-y’ -1 = 2y
$$
With $y=mathrme^-z$. Should be easier to solve.

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

share|cite|improve this answer

share|cite|improve this answer

answered Aug 19 at 11:51

Chinny84

11.8k21426

answered Aug 19 at 11:51

Chinny84

11.8k21426

answered Aug 19 at 11:51

Chinny84

11.8k21426

11.8k21426

add a comment | 

add a comment | 

up vote
1
down vote

it looks like Bernouilli 's equation
$$z'=e^z+2$$
substitute
$$u=e^z+2 implies frac dudz=e^z=u-2$$
The equation becomes
$$frac dzdt=u$$
$$frac dzdufrac dudt=u$$
$$frac dudt=ufrac dudz$$
$$frac dudt=u(u-2)$$
$$u'_t+2u=u^2$$
This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way
$$int frac duu(u-2)=int dt$$

share|cite|improve this answer

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

add a comment | 

up vote
1
down vote

it looks like Bernouilli 's equation
$$z'=e^z+2$$
substitute
$$u=e^z+2 implies frac dudz=e^z=u-2$$
The equation becomes
$$frac dzdt=u$$
$$frac dzdufrac dudt=u$$
$$frac dudt=ufrac dudz$$
$$frac dudt=u(u-2)$$
$$u'_t+2u=u^2$$
This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way
$$int frac duu(u-2)=int dt$$

share|cite|improve this answer

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

add a comment | 

up vote
1
down vote

up vote
1
down vote

it looks like Bernouilli 's equation
$$z'=e^z+2$$
substitute
$$u=e^z+2 implies frac dudz=e^z=u-2$$
The equation becomes
$$frac dzdt=u$$
$$frac dzdufrac dudt=u$$
$$frac dudt=ufrac dudz$$
$$frac dudt=u(u-2)$$
$$u'_t+2u=u^2$$
This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way
$$int frac duu(u-2)=int dt$$

share|cite|improve this answer

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

it looks like Bernouilli 's equation
$$z'=e^z+2$$
substitute
$$u=e^z+2 implies frac dudz=e^z=u-2$$
The equation becomes
$$frac dzdt=u$$
$$frac dzdufrac dudt=u$$
$$frac dudt=ufrac dudz$$
$$frac dudt=u(u-2)$$
$$u'_t+2u=u^2$$
This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way
$$int frac duu(u-2)=int dt$$

share|cite|improve this answer

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

share|cite|improve this answer

share|cite|improve this answer

edited Aug 19 at 12:47

edited Aug 19 at 12:47

edited Aug 19 at 12:47

answered Aug 19 at 12:38

Isham

10.9k3929

answered Aug 19 at 12:38

Isham

10.9k3929

answered Aug 19 at 12:38

Isham

10.9k3929

10.9k3929

add a comment | 

add a comment | 

 

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