How do I find the general solution for this ode












0












$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48
















0












$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48














0












0








0





$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 0:40







Steward

















asked Jan 3 at 0:33









StewardSteward

183




183












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48


















  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48
















$begingroup$
Yes, thank you, I will edit my post..
$endgroup$
– Steward
Jan 3 at 0:39




$begingroup$
Yes, thank you, I will edit my post..
$endgroup$
– Steward
Jan 3 at 0:39




1




1




$begingroup$
That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
$endgroup$
– DonAntonio
Jan 3 at 0:43






$begingroup$
That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
$endgroup$
– DonAntonio
Jan 3 at 0:43














$begingroup$
It was provided in the textbook it self ..
$endgroup$
– Steward
Jan 3 at 0:48




$begingroup$
It was provided in the textbook it self ..
$endgroup$
– Steward
Jan 3 at 0:48










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55
















4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55














4












4








4





$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$



Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 1:00

























answered Jan 3 at 0:37









John DoeJohn Doe

11.2k11239




11.2k11239












  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55


















  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55
















$begingroup$
Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
$endgroup$
– Steward
Jan 3 at 0:47






$begingroup$
Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
$endgroup$
– Steward
Jan 3 at 0:47














$begingroup$
@Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
$endgroup$
– DonAntonio
Jan 3 at 0:51




$begingroup$
@Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
$endgroup$
– DonAntonio
Jan 3 at 0:51












$begingroup$
@Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
$endgroup$
– John Doe
Jan 3 at 0:51




$begingroup$
@Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
$endgroup$
– John Doe
Jan 3 at 0:51












$begingroup$
In that case, what would be the correct answer ?
$endgroup$
– Steward
Jan 3 at 0:54




$begingroup$
In that case, what would be the correct answer ?
$endgroup$
– Steward
Jan 3 at 0:54












$begingroup$
@Steward see the edit.
$endgroup$
– John Doe
Jan 3 at 0:55




$begingroup$
@Steward see the edit.
$endgroup$
– John Doe
Jan 3 at 0:55


















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