Doubt in Section 1.5 of *Advanced Calculus for Applications* by Hildebrand
1
$begingroup$
enter image description here
I am learning differential equations from the book Advanced Calculus for Applications by Hildebrand but I am having difficulty in understanding what he is trying to tell in the section enclosed inside the blue bracket?
I do know $frac{partial[ L{e}^{rx}]}{partial r}=0 $ at $r=r_{1}$ but I cannot follow how this leads to the conclusion for giving the solution?
ordinary-differential-equations
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
$endgroup$
add a comment |
1
$begingroup$
enter image description here
I am learning differential equations from the book Advanced Calculus for Applications by Hildebrand but I am having difficulty in understanding what he is trying to tell in the section enclosed inside the blue bracket?
I do know $frac{partial[ L{e}^{rx}]}{partial r}=0 $ at $r=r_{1}$ but I cannot follow how this leads to the conclusion for giving the solution?
ordinary-differential-equations
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
$endgroup$
4
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03
add a comment |
1
1
1
0
$begingroup$
enter image description here
I am learning differential equations from the book Advanced Calculus for Applications by Hildebrand but I am having difficulty in understanding what he is trying to tell in the section enclosed inside the blue bracket?
I do know $frac{partial[ L{e}^{rx}]}{partial r}=0 $ at $r=r_{1}$ but I cannot follow how this leads to the conclusion for giving the solution?
ordinary-differential-equations
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
$endgroup$
enter image description here
I am learning differential equations from the book Advanced Calculus for Applications by Hildebrand but I am having difficulty in understanding what he is trying to tell in the section enclosed inside the blue bracket?
I do know $frac{partial[ L{e}^{rx}]}{partial r}=0 $ at $r=r_{1}$ but I cannot follow how this leads to the conclusion for giving the solution?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
edited Dec 10 '18 at 13:49
Brahadeesh
6,21242361
6,21242361
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
asked May 12 '18 at 12:12
curiosity zerocuriosity zero
84
84
4
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03
add a comment |
4
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03
4
4
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03
add a comment |
1 Answer
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active
oldest
votes
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If you differentiate the first expression in the blue box using the product rule and the chain rule the result will still have $(x-r_1)$ as a factor because $(x-r_1)$ occurs in the original with an exponent greater than $1$.
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
add a comment |
Your Answer
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1 Answer
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1 Answer
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1
$begingroup$
If you differentiate the first expression in the blue box using the product rule and the chain rule the result will still have $(x-r_1)$ as a factor because $(x-r_1)$ occurs in the original with an exponent greater than $1$.
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
add a comment |
1
$begingroup$
If you differentiate the first expression in the blue box using the product rule and the chain rule the result will still have $(x-r_1)$ as a factor because $(x-r_1)$ occurs in the original with an exponent greater than $1$.
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
add a comment |
1
1
1
$begingroup$
If you differentiate the first expression in the blue box using the product rule and the chain rule the result will still have $(x-r_1)$ as a factor because $(x-r_1)$ occurs in the original with an exponent greater than $1$.
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
If you differentiate the first expression in the blue box using the product rule and the chain rule the result will still have $(x-r_1)$ as a factor because $(x-r_1)$ occurs in the original with an exponent greater than $1$.
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
answered May 13 '18 at 12:28
Ethan BolkerEthan Bolker
42.4k549112
42.4k549112
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
add a comment |
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
I understand that part but I am having trouble following how this leads to the solution corresponding to the double root?
$endgroup$
– curiosity zero
May 13 '18 at 15:15
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
It works that way because of what happens when you differentiate $xe^{x}$. Just check that the last line really does give solutions to the homegeneous equation whatever the values of the constants.
$endgroup$
– Ethan Bolker
May 13 '18 at 15:33
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
$begingroup$
Ok, I finally understand now. Thank you for your attention.
$endgroup$
– curiosity zero
May 13 '18 at 15:59
add a comment |
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4
$begingroup$
Welcome to stackexchange. If you have really understood everything in the first $8$ pages then this explanation should be clear to you. I can't think of a way to explain it without using essentially the same argument. If you can point to the first particular sentence where you are stuck and say what you think it means perhaps we can help. Do that by editing the question, not in comments, and put the math in the question, not in an image. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
May 12 '18 at 12:17
$begingroup$
Thanks for posting the link . This is my first time asking a question on this site so I did know how equations were written. It was a great help. I have also included the part where I got struck more precisely.
$endgroup$
– curiosity zero
May 13 '18 at 8:03