Mixing
maximum interval of existence of a solution of differential equations
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have these initial value problems:
$x⋅u'(x) = u(x) + x^3⋅e^x^2$; $u(1)=e/2$
$u'(x) = 2xcdot u(x)^3$; $u(0) = a$; ($a in mathbb R$, hint: case differentiation)
$x^3⋅u'(x)-u(x)^3-x^2⋅u(x) = 0$; $u(1)=1$
I think, I already solved them correctly, with these solutions:
$u(x) = dfracxe^x^22$
$u(x) = dfrac1sqrt-2x^2+frac1a^2$
$u(x) = dfracxsqrt-2log(x)+1$
How can I find the maximum interval of existence now?
I hope y'all understand me, I'm not a native english speaker.
differential-equations analysis
edited Jul 16 at 8:55
Dylan
11.4k31026
asked Jul 15 at 12:49
Joshua
103
 |Â
show 2 more comments
up vote
0
down vote
favorite
I have these initial value problems:
$x⋅u'(x) = u(x) + x^3⋅e^x^2$; $u(1)=e/2$
$u'(x) = 2xcdot u(x)^3$; $u(0) = a$; ($a in mathbb R$, hint: case differentiation)
$x^3⋅u'(x)-u(x)^3-x^2⋅u(x) = 0$; $u(1)=1$
I think, I already solved them correctly, with these solutions:
$u(x) = dfracxe^x^22$
$u(x) = dfrac1sqrt-2x^2+frac1a^2$
$u(x) = dfracxsqrt-2log(x)+1$
How can I find the maximum interval of existence now?
I hope y'all understand me, I'm not a native english speaker.
differential-equations analysis
edited Jul 16 at 8:55
Dylan
11.4k31026
asked Jul 15 at 12:49
Joshua
103
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
Ah that's very kind
– Joshua
Jul 16 at 9:00
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have these initial value problems:
$x⋅u'(x) = u(x) + x^3⋅e^x^2$; $u(1)=e/2$
$u'(x) = 2xcdot u(x)^3$; $u(0) = a$; ($a in mathbb R$, hint: case differentiation)
$x^3⋅u'(x)-u(x)^3-x^2⋅u(x) = 0$; $u(1)=1$
I think, I already solved them correctly, with these solutions:
$u(x) = dfracxe^x^22$
$u(x) = dfrac1sqrt-2x^2+frac1a^2$
$u(x) = dfracxsqrt-2log(x)+1$
How can I find the maximum interval of existence now?
I hope y'all understand me, I'm not a native english speaker.
differential-equations analysis
edited Jul 16 at 8:55
Dylan
11.4k31026
asked Jul 15 at 12:49
Joshua
103
I have these initial value problems:
$x⋅u'(x) = u(x) + x^3⋅e^x^2$; $u(1)=e/2$
$u'(x) = 2xcdot u(x)^3$; $u(0) = a$; ($a in mathbb R$, hint: case differentiation)
$x^3⋅u'(x)-u(x)^3-x^2⋅u(x) = 0$; $u(1)=1$
I think, I already solved them correctly, with these solutions:
$u(x) = dfracxe^x^22$
$u(x) = dfrac1sqrt-2x^2+frac1a^2$
$u(x) = dfracxsqrt-2log(x)+1$
How can I find the maximum interval of existence now?
I hope y'all understand me, I'm not a native english speaker.
differential-equations analysis
edited Jul 16 at 8:55
Dylan
11.4k31026
asked Jul 15 at 12:49
Joshua
103
edited Jul 16 at 8:55
Dylan
11.4k31026
edited Jul 16 at 8:55
Dylan
11.4k31026
edited Jul 16 at 8:55
Dylan
11.4k31026
11.4k31026
asked Jul 15 at 12:49
Joshua
103
asked Jul 15 at 12:49
Joshua
103
asked Jul 15 at 12:49
Joshua
103
103
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
Ah that's very kind
– Joshua
Jul 16 at 9:00
 |Â
show 2 more comments
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
Ah that's very kind
– Joshua
Jul 16 at 9:00
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
Ah that's very kind
– Joshua
Jul 16 at 9:00
Ah that's very kind
– Joshua
Jul 16 at 9:00
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
The first DE equation is defined for $xneq 0$ and hence the solution is defined either in $(-infty, 0)$ or in $(0, infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, infty)$. Also the unique solution is $y=frac12x^2+frace-12x$.
For the second one consider two cases $a=0$ and $aneq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-infty, infty)$. For $aneq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $xneq 0$.
answered Jul 16 at 0:11
daulomb
3,4261617
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
add a comment |Â
up vote
1
down vote
I will briefly go through each problem, but it's up to you to fill in the details.
- Rewrite the equation in standard form
$$ u' -frac1xu = x^2e^x^2 $$
This is a standard first-order linear equation. The integrating factor is $mu = e^int -frac1xdx = e^-ln x = x^-1$. Multiplying through, we can simplify
$$ left(fracuxright)' = xe^x^2 implies fracux = frac12e^x^2 + c $$
The initial condition gives $c=0$, confirming your solution is correct.
As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-infty,0)$ or $(0,infty)$. The one that contains the given initial point is the answer here.
- Upon separating and integrating, you end up with the solution
$$ frac1u^2 = frac1a^2 - 2x^2 $$
Since the left hand side cannot be negative, a required condition is $2x^2 le frac1a^2$ or $|x| le frac1sqrt2$. This is your domain.
Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as
$$ u = fracasqrt1-2a^2x^2 $$
- This is a Bernoulli equation. The substitution $y=u^-2$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^-2 ge 0$, just like in problem 2.
answered Jul 16 at 9:20
Dylan
11.4k31026
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The first DE equation is defined for $xneq 0$ and hence the solution is defined either in $(-infty, 0)$ or in $(0, infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, infty)$. Also the unique solution is $y=frac12x^2+frace-12x$.
For the second one consider two cases $a=0$ and $aneq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-infty, infty)$. For $aneq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $xneq 0$.
answered Jul 16 at 0:11
daulomb
3,4261617
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
add a comment |Â
up vote
0
down vote
accepted
The first DE equation is defined for $xneq 0$ and hence the solution is defined either in $(-infty, 0)$ or in $(0, infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, infty)$. Also the unique solution is $y=frac12x^2+frace-12x$.
For the second one consider two cases $a=0$ and $aneq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-infty, infty)$. For $aneq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $xneq 0$.
answered Jul 16 at 0:11
daulomb
3,4261617
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The first DE equation is defined for $xneq 0$ and hence the solution is defined either in $(-infty, 0)$ or in $(0, infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, infty)$. Also the unique solution is $y=frac12x^2+frace-12x$.
For the second one consider two cases $a=0$ and $aneq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-infty, infty)$. For $aneq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $xneq 0$.
answered Jul 16 at 0:11
daulomb
3,4261617
The first DE equation is defined for $xneq 0$ and hence the solution is defined either in $(-infty, 0)$ or in $(0, infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, infty)$. Also the unique solution is $y=frac12x^2+frace-12x$.
For the second one consider two cases $a=0$ and $aneq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-infty, infty)$. For $aneq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $xneq 0$.
answered Jul 16 at 0:11
daulomb
3,4261617
answered Jul 16 at 0:11
daulomb
3,4261617
answered Jul 16 at 0:11
daulomb
3,4261617
answered Jul 16 at 0:11
daulomb
3,4261617
3,4261617
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
add a comment |Â
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
Thank you very much!!
– Joshua
Jul 16 at 8:41
Thank you very much!!
– Joshua
Jul 16 at 8:41
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
Why is the first one not defined for x=0?
– Joshua
Jul 16 at 9:09
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
because for this particular equation as the coefficient of $u'$, the leading term is $x$ which must be nonzero. So you should omit it from the definition of the solution. Also you can see what happens when you take $x=0$?
– daulomb
Jul 16 at 9:41
add a comment |Â
up vote
1
down vote
I will briefly go through each problem, but it's up to you to fill in the details.
- Rewrite the equation in standard form
$$ u' -frac1xu = x^2e^x^2 $$
This is a standard first-order linear equation. The integrating factor is $mu = e^int -frac1xdx = e^-ln x = x^-1$. Multiplying through, we can simplify
$$ left(fracuxright)' = xe^x^2 implies fracux = frac12e^x^2 + c $$
The initial condition gives $c=0$, confirming your solution is correct.
As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-infty,0)$ or $(0,infty)$. The one that contains the given initial point is the answer here.
- Upon separating and integrating, you end up with the solution
$$ frac1u^2 = frac1a^2 - 2x^2 $$
Since the left hand side cannot be negative, a required condition is $2x^2 le frac1a^2$ or $|x| le frac1sqrt2$. This is your domain.
Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as
$$ u = fracasqrt1-2a^2x^2 $$
- This is a Bernoulli equation. The substitution $y=u^-2$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^-2 ge 0$, just like in problem 2.
answered Jul 16 at 9:20
Dylan
11.4k31026
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
add a comment |Â
up vote
1
down vote
I will briefly go through each problem, but it's up to you to fill in the details.
- Rewrite the equation in standard form
$$ u' -frac1xu = x^2e^x^2 $$
This is a standard first-order linear equation. The integrating factor is $mu = e^int -frac1xdx = e^-ln x = x^-1$. Multiplying through, we can simplify
$$ left(fracuxright)' = xe^x^2 implies fracux = frac12e^x^2 + c $$
The initial condition gives $c=0$, confirming your solution is correct.
As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-infty,0)$ or $(0,infty)$. The one that contains the given initial point is the answer here.
- Upon separating and integrating, you end up with the solution
$$ frac1u^2 = frac1a^2 - 2x^2 $$
Since the left hand side cannot be negative, a required condition is $2x^2 le frac1a^2$ or $|x| le frac1sqrt2$. This is your domain.
Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as
$$ u = fracasqrt1-2a^2x^2 $$
- This is a Bernoulli equation. The substitution $y=u^-2$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^-2 ge 0$, just like in problem 2.
answered Jul 16 at 9:20
Dylan
11.4k31026
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I will briefly go through each problem, but it's up to you to fill in the details.
- Rewrite the equation in standard form
$$ u' -frac1xu = x^2e^x^2 $$
This is a standard first-order linear equation. The integrating factor is $mu = e^int -frac1xdx = e^-ln x = x^-1$. Multiplying through, we can simplify
$$ left(fracuxright)' = xe^x^2 implies fracux = frac12e^x^2 + c $$
The initial condition gives $c=0$, confirming your solution is correct.
As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-infty,0)$ or $(0,infty)$. The one that contains the given initial point is the answer here.
- Upon separating and integrating, you end up with the solution
$$ frac1u^2 = frac1a^2 - 2x^2 $$
Since the left hand side cannot be negative, a required condition is $2x^2 le frac1a^2$ or $|x| le frac1sqrt2$. This is your domain.
Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as
$$ u = fracasqrt1-2a^2x^2 $$
- This is a Bernoulli equation. The substitution $y=u^-2$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^-2 ge 0$, just like in problem 2.
answered Jul 16 at 9:20
Dylan
11.4k31026
I will briefly go through each problem, but it's up to you to fill in the details.
- Rewrite the equation in standard form
$$ u' -frac1xu = x^2e^x^2 $$
This is a standard first-order linear equation. The integrating factor is $mu = e^int -frac1xdx = e^-ln x = x^-1$. Multiplying through, we can simplify
$$ left(fracuxright)' = xe^x^2 implies fracux = frac12e^x^2 + c $$
The initial condition gives $c=0$, confirming your solution is correct.
As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-infty,0)$ or $(0,infty)$. The one that contains the given initial point is the answer here.
- Upon separating and integrating, you end up with the solution
$$ frac1u^2 = frac1a^2 - 2x^2 $$
Since the left hand side cannot be negative, a required condition is $2x^2 le frac1a^2$ or $|x| le frac1sqrt2$. This is your domain.
Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as
$$ u = fracasqrt1-2a^2x^2 $$
- This is a Bernoulli equation. The substitution $y=u^-2$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^-2 ge 0$, just like in problem 2.
answered Jul 16 at 9:20
Dylan
11.4k31026
edited Jul 16 at 9:52
answered Jul 16 at 9:20
Dylan
11.4k31026
answered Jul 16 at 9:20
Dylan
11.4k31026
answered Jul 16 at 9:20
Dylan
11.4k31026
11.4k31026
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
add a comment |Â
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
Thank you very much for that detailed answer, seems like I have a big lack of knowledge, cause I'm not really able to follow your first calculation :/
– Joshua
Jul 16 at 9:36
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
For the first problem? What did you do to obtain the solution then?
– Dylan
Jul 16 at 9:52
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
Sorry for the late answer. I was going to send a Foto of my calculation, but that didn't work. However we just talked about the exercises in class and the teacher said, the domain of existence of the first equation is R
– Joshua
Jul 19 at 14:06
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%2f2852496%2fmaximum-interval-of-existence-of-a-solution-of-differential-equations%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
Can you show your work for these solutions? I don't think the first one is right
– Dylan
Jul 15 at 17:41
How can I send pics here? I have it on paper
– Joshua
Jul 15 at 21:29
Oh wait the fraction line should be under the whole pruduct including e, then it's correct right?
– Joshua
Jul 15 at 21:32
I fixed the fraction for you. Your initial post was a bit ambiguous, so I don't blame the editor. Here's a quick tutorial for the math formatting on this site
– Dylan
Jul 16 at 8:57
Ah that's very kind
– Joshua
Jul 16 at 9:00