Solve $(1-x^2)y''-xy'-y=e^{arcsin x}$. [closed]
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Problem
Solve $$(1-x^2)y''-xy'-y=e^{arcsin x}.$$
Can it be solved to get the general solution?
ordinary-differential-equations
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
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closed as off-topic by Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0 Dec 17 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Problem
Solve $$(1-x^2)y''-xy'-y=e^{arcsin x}.$$
Can it be solved to get the general solution?
ordinary-differential-equations
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
$endgroup$
closed as off-topic by Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0 Dec 17 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
4
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What have you tried?
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– Christoph
Dec 17 '18 at 12:07
1
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You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
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– Ben W
Dec 17 '18 at 12:17
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Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
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– Claude Leibovici
Dec 17 '18 at 12:34
add a comment |
$begingroup$
Problem
Solve $$(1-x^2)y''-xy'-y=e^{arcsin x}.$$
Can it be solved to get the general solution?
ordinary-differential-equations
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
$endgroup$
Problem
Solve $$(1-x^2)y''-xy'-y=e^{arcsin x}.$$
Can it be solved to get the general solution?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
edited Dec 17 '18 at 12:06
mengdie1982
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
asked Dec 17 '18 at 12:01
mengdie1982mengdie1982
4,927618
4,927618
closed as off-topic by Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0 Dec 17 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0 Dec 17 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Christoph, Riccardo.Alestra, user10354138, Adrian Keister, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
What have you tried?
$endgroup$
– Christoph
Dec 17 '18 at 12:07
1
$begingroup$
You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
$endgroup$
– Ben W
Dec 17 '18 at 12:17
$begingroup$
Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
$endgroup$
– Claude Leibovici
Dec 17 '18 at 12:34
add a comment |
4
$begingroup$
What have you tried?
$endgroup$
– Christoph
Dec 17 '18 at 12:07
1
$begingroup$
You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
$endgroup$
– Ben W
Dec 17 '18 at 12:17
$begingroup$
Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
$endgroup$
– Claude Leibovici
Dec 17 '18 at 12:34
4
4
$begingroup$
What have you tried?
$endgroup$
– Christoph
Dec 17 '18 at 12:07
$begingroup$
What have you tried?
$endgroup$
– Christoph
Dec 17 '18 at 12:07
1
1
$begingroup$
You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
$endgroup$
– Ben W
Dec 17 '18 at 12:17
$begingroup$
You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
$endgroup$
– Ben W
Dec 17 '18 at 12:17
$begingroup$
Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
$endgroup$
– Claude Leibovici
Dec 17 '18 at 12:34
$begingroup$
Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
$endgroup$
– Claude Leibovici
Dec 17 '18 at 12:34
add a comment |
1 Answer
1
active
oldest
votes
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Let $t=sin^{-1}x$ ,
Then $x=sin t$
$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=(sec t)dfrac{dy}{dt}$
$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left((sec t)dfrac{dy}{dt}right)=dfrac{dfrac{d}{dt}left((sec t)dfrac{dy}{dt}right)}{dfrac{dx}{dt}}=dfrac{(sec t)dfrac{d^2y}{dt^2}+(sec ttan t)dfrac{dy}{dt}}{cos t}=(sec^2t)dfrac{d^2y}{dt^2}+(sec^2ttan t)dfrac{dy}{dt}$
$thereforedfrac{d^2y}{dt^2}+(tan t)dfrac{dy}{dt}-(tan t)dfrac{dy}{dt}-y=e^t$
$dfrac{d^2y}{dt^2}-y=e^t$
$y=C_1e^t+C_2e^{-t}+dfrac{te^t}{2}$
$y=C_1e^{sin^{-1}x}+C_2e^{-sin^{-1}x}+dfrac{e^{sin^{-1}x}sin^{-1}x}{2}$
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $t=sin^{-1}x$ ,
Then $x=sin t$
$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=(sec t)dfrac{dy}{dt}$
$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left((sec t)dfrac{dy}{dt}right)=dfrac{dfrac{d}{dt}left((sec t)dfrac{dy}{dt}right)}{dfrac{dx}{dt}}=dfrac{(sec t)dfrac{d^2y}{dt^2}+(sec ttan t)dfrac{dy}{dt}}{cos t}=(sec^2t)dfrac{d^2y}{dt^2}+(sec^2ttan t)dfrac{dy}{dt}$
$thereforedfrac{d^2y}{dt^2}+(tan t)dfrac{dy}{dt}-(tan t)dfrac{dy}{dt}-y=e^t$
$dfrac{d^2y}{dt^2}-y=e^t$
$y=C_1e^t+C_2e^{-t}+dfrac{te^t}{2}$
$y=C_1e^{sin^{-1}x}+C_2e^{-sin^{-1}x}+dfrac{e^{sin^{-1}x}sin^{-1}x}{2}$
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
add a comment |
$begingroup$
Let $t=sin^{-1}x$ ,
Then $x=sin t$
$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=(sec t)dfrac{dy}{dt}$
$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left((sec t)dfrac{dy}{dt}right)=dfrac{dfrac{d}{dt}left((sec t)dfrac{dy}{dt}right)}{dfrac{dx}{dt}}=dfrac{(sec t)dfrac{d^2y}{dt^2}+(sec ttan t)dfrac{dy}{dt}}{cos t}=(sec^2t)dfrac{d^2y}{dt^2}+(sec^2ttan t)dfrac{dy}{dt}$
$thereforedfrac{d^2y}{dt^2}+(tan t)dfrac{dy}{dt}-(tan t)dfrac{dy}{dt}-y=e^t$
$dfrac{d^2y}{dt^2}-y=e^t$
$y=C_1e^t+C_2e^{-t}+dfrac{te^t}{2}$
$y=C_1e^{sin^{-1}x}+C_2e^{-sin^{-1}x}+dfrac{e^{sin^{-1}x}sin^{-1}x}{2}$
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
add a comment |
$begingroup$
Let $t=sin^{-1}x$ ,
Then $x=sin t$
$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=(sec t)dfrac{dy}{dt}$
$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left((sec t)dfrac{dy}{dt}right)=dfrac{dfrac{d}{dt}left((sec t)dfrac{dy}{dt}right)}{dfrac{dx}{dt}}=dfrac{(sec t)dfrac{d^2y}{dt^2}+(sec ttan t)dfrac{dy}{dt}}{cos t}=(sec^2t)dfrac{d^2y}{dt^2}+(sec^2ttan t)dfrac{dy}{dt}$
$thereforedfrac{d^2y}{dt^2}+(tan t)dfrac{dy}{dt}-(tan t)dfrac{dy}{dt}-y=e^t$
$dfrac{d^2y}{dt^2}-y=e^t$
$y=C_1e^t+C_2e^{-t}+dfrac{te^t}{2}$
$y=C_1e^{sin^{-1}x}+C_2e^{-sin^{-1}x}+dfrac{e^{sin^{-1}x}sin^{-1}x}{2}$
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
Let $t=sin^{-1}x$ ,
Then $x=sin t$
$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=(sec t)dfrac{dy}{dt}$
$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left((sec t)dfrac{dy}{dt}right)=dfrac{dfrac{d}{dt}left((sec t)dfrac{dy}{dt}right)}{dfrac{dx}{dt}}=dfrac{(sec t)dfrac{d^2y}{dt^2}+(sec ttan t)dfrac{dy}{dt}}{cos t}=(sec^2t)dfrac{d^2y}{dt^2}+(sec^2ttan t)dfrac{dy}{dt}$
$thereforedfrac{d^2y}{dt^2}+(tan t)dfrac{dy}{dt}-(tan t)dfrac{dy}{dt}-y=e^t$
$dfrac{d^2y}{dt^2}-y=e^t$
$y=C_1e^t+C_2e^{-t}+dfrac{te^t}{2}$
$y=C_1e^{sin^{-1}x}+C_2e^{-sin^{-1}x}+dfrac{e^{sin^{-1}x}sin^{-1}x}{2}$
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
answered Dec 17 '18 at 12:46
doraemonpauldoraemonpaul
12.6k31660
12.6k31660
add a comment |
add a comment |
4
$begingroup$
What have you tried?
$endgroup$
– Christoph
Dec 17 '18 at 12:07
1
$begingroup$
You have rep of over 4000. I have to assume there is a language difficulty here, because this is a very poorly worded question.
$endgroup$
– Ben W
Dec 17 '18 at 12:17
$begingroup$
Did you solve at least $(1-x^2)y''-xy'-y=0$ ?
$endgroup$
– Claude Leibovici
Dec 17 '18 at 12:34