Mixing
ODE $ x(y+4)+fracdydx=0 $, with conditions leading to a log of negative number?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Trivial ODE and trivial question:
$$ x(y+4)+fracdydx=0 $$ with initial conditions $y=-5, x=0$
After we separate variables we get:
$$ -fracdyy+4=x,dx $$
Integrate left and write parts:
$$-ln(y+4)=frac12x^2 + C $$
Here we see that if $y=-5 $ we have a log of negative number.
Trying not to think about it, I proceed as follows:
Exponentiation of both parts:
$$y = e^-1/2x^2-C - 4$$
And then:
$$y = e^-1/2x^2e^-C - 4$$
$$y = e^-1/2x^2C - 4$$
(this new $C $ to $e^-C$ and cannot be negative)
$$-5 = e^0C - 4$$
$$C=-1$$
(but we see that it is in fact negative under given initial conditions)
which gives us the correct answer:
$$y = -e^-1/2x^2 - 4$$
Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative $C $ (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?
Thanks!
differential-equations
edited Jul 26 at 16:32
amWhy
189k25219431
asked Jul 23 at 21:43
type2
1285
 |Â
show 1 more comment
up vote
3
down vote
favorite
Trivial ODE and trivial question:
$$ x(y+4)+fracdydx=0 $$ with initial conditions $y=-5, x=0$
After we separate variables we get:
$$ -fracdyy+4=x,dx $$
Integrate left and write parts:
$$-ln(y+4)=frac12x^2 + C $$
Here we see that if $y=-5 $ we have a log of negative number.
Trying not to think about it, I proceed as follows:
Exponentiation of both parts:
$$y = e^-1/2x^2-C - 4$$
And then:
$$y = e^-1/2x^2e^-C - 4$$
$$y = e^-1/2x^2C - 4$$
(this new $C $ to $e^-C$ and cannot be negative)
$$-5 = e^0C - 4$$
$$C=-1$$
(but we see that it is in fact negative under given initial conditions)
which gives us the correct answer:
$$y = -e^-1/2x^2 - 4$$
Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative $C $ (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?
Thanks!
differential-equations
edited Jul 26 at 16:32
amWhy
189k25219431
asked Jul 23 at 21:43
type2
1285
4
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
That would be a good idea!
– amWhy
Jul 23 at 22:01
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Trivial ODE and trivial question:
$$ x(y+4)+fracdydx=0 $$ with initial conditions $y=-5, x=0$
After we separate variables we get:
$$ -fracdyy+4=x,dx $$
Integrate left and write parts:
$$-ln(y+4)=frac12x^2 + C $$
Here we see that if $y=-5 $ we have a log of negative number.
Trying not to think about it, I proceed as follows:
Exponentiation of both parts:
$$y = e^-1/2x^2-C - 4$$
And then:
$$y = e^-1/2x^2e^-C - 4$$
$$y = e^-1/2x^2C - 4$$
(this new $C $ to $e^-C$ and cannot be negative)
$$-5 = e^0C - 4$$
$$C=-1$$
(but we see that it is in fact negative under given initial conditions)
which gives us the correct answer:
$$y = -e^-1/2x^2 - 4$$
Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative $C $ (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?
Thanks!
differential-equations
edited Jul 26 at 16:32
amWhy
189k25219431
asked Jul 23 at 21:43
type2
1285
Trivial ODE and trivial question:
$$ x(y+4)+fracdydx=0 $$ with initial conditions $y=-5, x=0$
After we separate variables we get:
$$ -fracdyy+4=x,dx $$
Integrate left and write parts:
$$-ln(y+4)=frac12x^2 + C $$
Here we see that if $y=-5 $ we have a log of negative number.
Trying not to think about it, I proceed as follows:
Exponentiation of both parts:
$$y = e^-1/2x^2-C - 4$$
And then:
$$y = e^-1/2x^2e^-C - 4$$
$$y = e^-1/2x^2C - 4$$
(this new $C $ to $e^-C$ and cannot be negative)
$$-5 = e^0C - 4$$
$$C=-1$$
(but we see that it is in fact negative under given initial conditions)
which gives us the correct answer:
$$y = -e^-1/2x^2 - 4$$
Now I feel that I've cheated somewhere. My only guess that I was able to deal with a log of a negative number and get negative $C $ (which shouldn't be negative) is because I have involved complex numbers between the lines. Is it so? If not, how can I make this solution rigorous?
Thanks!
differential-equations
edited Jul 26 at 16:32
amWhy
189k25219431
asked Jul 23 at 21:43
type2
1285
edited Jul 26 at 16:32
amWhy
189k25219431
edited Jul 26 at 16:32
amWhy
189k25219431
edited Jul 26 at 16:32
amWhy
189k25219431
189k25219431
asked Jul 23 at 21:43
type2
1285
asked Jul 23 at 21:43
type2
1285
asked Jul 23 at 21:43
type2
1285
1285
4
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
That would be a good idea!
– amWhy
Jul 23 at 22:01
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18
 |Â
show 1 more comment
4
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
That would be a good idea!
– amWhy
Jul 23 at 22:01
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18
4
4
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
That would be a good idea!
– amWhy
Jul 23 at 22:01
That would be a good idea!
– amWhy
Jul 23 at 22:01
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Re: your first integration: remember that $$int fracf'(y)f(y)dy = ln(|f(y)|) + C$$
So in your case, the LHS should result in $−ln(|y+4|).$ Hence ...
answered Jul 23 at 22:24
amWhy
189k25219431
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Re: your first integration: remember that $$int fracf'(y)f(y)dy = ln(|f(y)|) + C$$
So in your case, the LHS should result in $−ln(|y+4|).$ Hence ...
answered Jul 23 at 22:24
amWhy
189k25219431
add a comment |Â
up vote
3
down vote
accepted
Hint:
Re: your first integration: remember that $$int fracf'(y)f(y)dy = ln(|f(y)|) + C$$
So in your case, the LHS should result in $−ln(|y+4|).$ Hence ...
answered Jul 23 at 22:24
amWhy
189k25219431
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint:
Re: your first integration: remember that $$int fracf'(y)f(y)dy = ln(|f(y)|) + C$$
So in your case, the LHS should result in $−ln(|y+4|).$ Hence ...
answered Jul 23 at 22:24
amWhy
189k25219431
Hint:
Re: your first integration: remember that $$int fracf'(y)f(y)dy = ln(|f(y)|) + C$$
So in your case, the LHS should result in $−ln(|y+4|).$ Hence ...
answered Jul 23 at 22:24
amWhy
189k25219431
answered Jul 23 at 22:24
amWhy
189k25219431
answered Jul 23 at 22:24
amWhy
189k25219431
answered Jul 23 at 22:24
amWhy
189k25219431
189k25219431
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860803%2fode-xy4-fracdydx-0-with-conditions-leading-to-a-log-of-negative-nu%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
Re your first integration, remember that $int fracf'(y)f(y) dy$ on the left, is $ln(|f(y)| + C$. So in your case, the LHS should result in $-ln(|y+4|)$. Hence ...
– amWhy
Jul 23 at 21:47
@amWhy Thanks, I totally forgot about it. Then $ |y+4| = e^-1/2x^2-C $. Should I get rid of abs value by considering $ y<-4 $ and $ y ge-4 $ next ?
– type2
Jul 23 at 21:58
That would be a good idea!
– amWhy
Jul 23 at 22:01
@amWhy So if $ y<-4 $: $ -y-4=e^-1/2x^2-C $ and $ y= -e^-1/2x^2-C - 4$, then $ -5 = -e^-C - 4 $ and hence $ C=0 $ which gives the correct answer. Now if if $ y ge -4 $: $ y+4=e^-1/2x^2-C $ and and $ y= e^-1/2x^2-C - 4$ then $ -5= e^-C - 4$ which has no real solutions for $ C $, right?
– type2
Jul 23 at 22:16
@amWhy btw, feel free to post your first comment as an answer and I will accept it, thanks!
– type2
Jul 23 at 22:18