How to solve this kind of nonlinear differential equation?
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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
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
asked Aug 19 at 11:46
MysteryGuy
377216
add a comment |Â
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
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
asked Aug 19 at 11:46
MysteryGuy
377216
add a comment |Â
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
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
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
differential-equations
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
asked Aug 19 at 11:46
MysteryGuy
377216
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
asked Aug 19 at 11:46
MysteryGuy
377216
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
edited Aug 19 at 11:52
Rodrigo de Azevedo
12.7k41752
12.7k41752
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 |Â
add a comment |Â
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.
answered Aug 19 at 11:51
Chinny84
11.9k21426
add a comment |Â
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
$$int frac duu(u-2)=int dt$$
answered Aug 19 at 12:38
Isham
11.1k3929
add a comment |Â
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.
answered Aug 19 at 11:51
Chinny84
11.9k21426
add a comment |Â
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.
answered Aug 19 at 11:51
Chinny84
11.9k21426
add a comment |Â
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.
answered Aug 19 at 11:51
Chinny84
11.9k21426
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.
answered Aug 19 at 11:51
Chinny84
11.9k21426
answered Aug 19 at 11:51
Chinny84
11.9k21426
answered Aug 19 at 11:51
Chinny84
11.9k21426
answered Aug 19 at 11:51
Chinny84
11.9k21426
11.9k21426
add a comment |Â
add a comment |Â
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
$$int frac duu(u-2)=int dt$$
answered Aug 19 at 12:38
Isham
11.1k3929
add a comment |Â
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
$$int frac duu(u-2)=int dt$$
answered Aug 19 at 12:38
Isham
11.1k3929
add a comment |Â
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
$$int frac duu(u-2)=int dt$$
answered Aug 19 at 12:38
Isham
11.1k3929
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
$$int frac duu(u-2)=int dt$$
answered Aug 19 at 12:38
Isham
11.1k3929
edited Aug 19 at 12:47
answered Aug 19 at 12:38
Isham
11.1k3929
answered Aug 19 at 12:38
Isham
11.1k3929
answered Aug 19 at 12:38
Isham
11.1k3929
11.1k3929
add a comment |Â
add a comment |Â
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