Confusion with how to handle constant in differential equation











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I'm trying to learn to solve differential equations. Currently in class we are discussing integration factors. Here is one of the problems on the homework -



$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$



We can quickly see that if we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term and we see that they are not equal, so we need an integrating factor.



An integrating factor of $x$ gets us a bit farther:



$$(3x^2e^y+2xy)dx + (x^3e^y+x^2)dy=0tag{2}$$



And now when we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term , the two are equal.



From here, I think we would integrate the left term with respect to $x$ and right term with respect to $y$ as follows -



$$int {(3x^2e^y+2xy)dx} + int (x^3e^y+x^2)dy = 0 tag{3}$$
$$2x^3e^y+2x^2y+C=0tag{4}$$
$$x^3e^y+x^2y=Ctag{5}$$



Am I thinking about this the right way?



Also, in $(5)$, is it fine to move the Constant to the right and absorb the common coefficient of 2 into it to simplify the expression?
$(5)$ is the answer in the book and (4) was my answer. Either I have done something wrong, or I am confused in regard to what may be combined into the constant $C$.










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  • 1




    Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
    – Isham
    2 days ago










  • Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
    – Claude Leibovici
    2 days ago















up vote
1
down vote

favorite












I'm trying to learn to solve differential equations. Currently in class we are discussing integration factors. Here is one of the problems on the homework -



$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$



We can quickly see that if we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term and we see that they are not equal, so we need an integrating factor.



An integrating factor of $x$ gets us a bit farther:



$$(3x^2e^y+2xy)dx + (x^3e^y+x^2)dy=0tag{2}$$



And now when we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term , the two are equal.



From here, I think we would integrate the left term with respect to $x$ and right term with respect to $y$ as follows -



$$int {(3x^2e^y+2xy)dx} + int (x^3e^y+x^2)dy = 0 tag{3}$$
$$2x^3e^y+2x^2y+C=0tag{4}$$
$$x^3e^y+x^2y=Ctag{5}$$



Am I thinking about this the right way?



Also, in $(5)$, is it fine to move the Constant to the right and absorb the common coefficient of 2 into it to simplify the expression?
$(5)$ is the answer in the book and (4) was my answer. Either I have done something wrong, or I am confused in regard to what may be combined into the constant $C$.










share|cite|improve this question




















  • 1




    Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
    – Isham
    2 days ago










  • Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
    – Claude Leibovici
    2 days ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to learn to solve differential equations. Currently in class we are discussing integration factors. Here is one of the problems on the homework -



$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$



We can quickly see that if we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term and we see that they are not equal, so we need an integrating factor.



An integrating factor of $x$ gets us a bit farther:



$$(3x^2e^y+2xy)dx + (x^3e^y+x^2)dy=0tag{2}$$



And now when we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term , the two are equal.



From here, I think we would integrate the left term with respect to $x$ and right term with respect to $y$ as follows -



$$int {(3x^2e^y+2xy)dx} + int (x^3e^y+x^2)dy = 0 tag{3}$$
$$2x^3e^y+2x^2y+C=0tag{4}$$
$$x^3e^y+x^2y=Ctag{5}$$



Am I thinking about this the right way?



Also, in $(5)$, is it fine to move the Constant to the right and absorb the common coefficient of 2 into it to simplify the expression?
$(5)$ is the answer in the book and (4) was my answer. Either I have done something wrong, or I am confused in regard to what may be combined into the constant $C$.










share|cite|improve this question















I'm trying to learn to solve differential equations. Currently in class we are discussing integration factors. Here is one of the problems on the homework -



$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$



We can quickly see that if we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term and we see that they are not equal, so we need an integrating factor.



An integrating factor of $x$ gets us a bit farther:



$$(3x^2e^y+2xy)dx + (x^3e^y+x^2)dy=0tag{2}$$



And now when we take $frac{partial}{partial y}$ of the left term and $frac{partial}{partial x}$ of the right term , the two are equal.



From here, I think we would integrate the left term with respect to $x$ and right term with respect to $y$ as follows -



$$int {(3x^2e^y+2xy)dx} + int (x^3e^y+x^2)dy = 0 tag{3}$$
$$2x^3e^y+2x^2y+C=0tag{4}$$
$$x^3e^y+x^2y=Ctag{5}$$



Am I thinking about this the right way?



Also, in $(5)$, is it fine to move the Constant to the right and absorb the common coefficient of 2 into it to simplify the expression?
$(5)$ is the answer in the book and (4) was my answer. Either I have done something wrong, or I am confused in regard to what may be combined into the constant $C$.







integration differential-equations






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share|cite|improve this question













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edited 2 days ago

























asked 2 days ago









blueether

1134




1134








  • 1




    Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
    – Isham
    2 days ago










  • Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
    – Claude Leibovici
    2 days ago














  • 1




    Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
    – Isham
    2 days ago










  • Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
    – Claude Leibovici
    2 days ago








1




1




Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
– Isham
2 days ago




Here you have some examples with exercices and full solutions. You may find it helpful tutorial.math.lamar.edu/Classes/DE/Exact.aspx
– Isham
2 days ago












Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
– Claude Leibovici
2 days ago




Just for your curiosity, there is an analytical expression of $y(x)$ in terms of Lambert function.
– Claude Leibovici
2 days ago










1 Answer
1






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oldest

votes

















up vote
0
down vote



accepted










$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$
multiply by X as integrating factor as you wrote
$$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$
Rearrange terms
$$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$
$$d(x^3e^y)+d(x^2y)=0$$
After integration
$$x^3e^y+x^2y=K$$
For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote...
$$2x^3e^y+2x^2y+C=0$$
$$implies x^3e^y+x^2y=-frac C 2$$
Substitute
$$K=-frac C2 implies x^3e^y+x^2y=K$$
K is just a constant.






share|cite|improve this answer























  • Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
    – blueether
    2 days ago










  • I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
    – Isham
    2 days ago








  • 1




    Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
    – blueether
    2 days ago






  • 1




    Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
    – Isham
    2 days ago













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1 Answer
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1 Answer
1






active

oldest

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oldest

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active

oldest

votes








up vote
0
down vote



accepted










$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$
multiply by X as integrating factor as you wrote
$$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$
Rearrange terms
$$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$
$$d(x^3e^y)+d(x^2y)=0$$
After integration
$$x^3e^y+x^2y=K$$
For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote...
$$2x^3e^y+2x^2y+C=0$$
$$implies x^3e^y+x^2y=-frac C 2$$
Substitute
$$K=-frac C2 implies x^3e^y+x^2y=K$$
K is just a constant.






share|cite|improve this answer























  • Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
    – blueether
    2 days ago










  • I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
    – Isham
    2 days ago








  • 1




    Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
    – blueether
    2 days ago






  • 1




    Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
    – Isham
    2 days ago

















up vote
0
down vote



accepted










$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$
multiply by X as integrating factor as you wrote
$$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$
Rearrange terms
$$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$
$$d(x^3e^y)+d(x^2y)=0$$
After integration
$$x^3e^y+x^2y=K$$
For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote...
$$2x^3e^y+2x^2y+C=0$$
$$implies x^3e^y+x^2y=-frac C 2$$
Substitute
$$K=-frac C2 implies x^3e^y+x^2y=K$$
K is just a constant.






share|cite|improve this answer























  • Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
    – blueether
    2 days ago










  • I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
    – Isham
    2 days ago








  • 1




    Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
    – blueether
    2 days ago






  • 1




    Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
    – Isham
    2 days ago















up vote
0
down vote



accepted







up vote
0
down vote



accepted






$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$
multiply by X as integrating factor as you wrote
$$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$
Rearrange terms
$$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$
$$d(x^3e^y)+d(x^2y)=0$$
After integration
$$x^3e^y+x^2y=K$$
For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote...
$$2x^3e^y+2x^2y+C=0$$
$$implies x^3e^y+x^2y=-frac C 2$$
Substitute
$$K=-frac C2 implies x^3e^y+x^2y=K$$
K is just a constant.






share|cite|improve this answer














$$(3xe^y+2y)dx + (x^2e^y+x)dy=0tag{1}$$
multiply by X as integrating factor as you wrote
$$(3x^2e^y+2yx)dx + (x^3e^y+x^2)dy=0$$
Rearrange terms
$$(3x^2e^ydx+x^3e^ydy)+(2yxdx + x^2dy)=0$$
$$d(x^3e^y)+d(x^2y)=0$$
After integration
$$x^3e^y+x^2y=K$$
For the constant yes it absorbs the factor 2 and you can have it on the other side. Your result is correct but I wouldn't write the integrals you wrote...
$$2x^3e^y+2x^2y+C=0$$
$$implies x^3e^y+x^2y=-frac C 2$$
Substitute
$$K=-frac C2 implies x^3e^y+x^2y=K$$
K is just a constant.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Isham

12.6k3929




12.6k3929












  • Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
    – blueether
    2 days ago










  • I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
    – Isham
    2 days ago








  • 1




    Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
    – blueether
    2 days ago






  • 1




    Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
    – Isham
    2 days ago




















  • Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
    – blueether
    2 days ago










  • I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
    – Isham
    2 days ago








  • 1




    Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
    – blueether
    2 days ago






  • 1




    Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
    – Isham
    2 days ago


















Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
– blueether
2 days ago




Thank you, this is helpful. Could you expand on why you wouldn't write the integrals that I wrote?
– blueether
2 days ago












I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
– Isham
2 days ago






I used to write them separeately not in an single equation..@blueether How can you justify the step from equation 2 to 3 ?? But if the teacher allows you to do that ...thats it
– Isham
2 days ago






1




1




Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
– blueether
2 days ago




Ah, I see. I've been away from the math for a decade or so and trying to get back into it. I was figuring, since the equation is equal to zero I could move one of the terms to the other side, then integrate both sides with respect to x on the left and y on the right (I simply didn't bother moving anything around). I appreciate your feedback, being an adult and back in school I am far less interested in what the teacher allows and would rather learn to do things the proper way. Style is certainly important.
– blueether
2 days ago




1




1




Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
– Isham
2 days ago






Well you did a good job except for that step 2 to 3...@blueether note that when you integrate a function of two variable with respect to one variable then the constant of integration is a function of the other variable..For the constant as you can see you made ,no mistake and your answer is the same as that of your book Call the constant C or K these are just names thats all
– Isham
2 days ago




















 

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