Find the general solution of $y(x)$ in $xe^{2y}y'-1-e^{2y}=0$












1












$begingroup$



Find the general solution of $y(x)$ in $xe^{2y}y'-1-e^{2y}=0$.




I'm just starting with diff equations and I'm having a really hard time solving them.



Here's what I've been doing:



I re-wrote the equation as:



$xe^{2y}frac{dy}{dx}-1-e^{2y}=0$, and then I ordered it as



$frac{1}{x}dx$=$frac{e^{2y}}{1+e^{2y}}dy$.



And I found the integral of both sides:



$int frac{1}{x}dx = int frac{e^{2y}}{1+e^{2y}}dy $



$ln(x)=frac{ln(e^{2y}+1)}{2}+c$.



I re-order it, and I still can't get any of the solutions... What I'm doing wrong?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do the possible solutions look like?
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:29










  • $begingroup$
    You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:32


















1












$begingroup$



Find the general solution of $y(x)$ in $xe^{2y}y'-1-e^{2y}=0$.




I'm just starting with diff equations and I'm having a really hard time solving them.



Here's what I've been doing:



I re-wrote the equation as:



$xe^{2y}frac{dy}{dx}-1-e^{2y}=0$, and then I ordered it as



$frac{1}{x}dx$=$frac{e^{2y}}{1+e^{2y}}dy$.



And I found the integral of both sides:



$int frac{1}{x}dx = int frac{e^{2y}}{1+e^{2y}}dy $



$ln(x)=frac{ln(e^{2y}+1)}{2}+c$.



I re-order it, and I still can't get any of the solutions... What I'm doing wrong?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do the possible solutions look like?
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:29










  • $begingroup$
    You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:32
















1












1








1





$begingroup$



Find the general solution of $y(x)$ in $xe^{2y}y'-1-e^{2y}=0$.




I'm just starting with diff equations and I'm having a really hard time solving them.



Here's what I've been doing:



I re-wrote the equation as:



$xe^{2y}frac{dy}{dx}-1-e^{2y}=0$, and then I ordered it as



$frac{1}{x}dx$=$frac{e^{2y}}{1+e^{2y}}dy$.



And I found the integral of both sides:



$int frac{1}{x}dx = int frac{e^{2y}}{1+e^{2y}}dy $



$ln(x)=frac{ln(e^{2y}+1)}{2}+c$.



I re-order it, and I still can't get any of the solutions... What I'm doing wrong?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$





Find the general solution of $y(x)$ in $xe^{2y}y'-1-e^{2y}=0$.




I'm just starting with diff equations and I'm having a really hard time solving them.



Here's what I've been doing:



I re-wrote the equation as:



$xe^{2y}frac{dy}{dx}-1-e^{2y}=0$, and then I ordered it as



$frac{1}{x}dx$=$frac{e^{2y}}{1+e^{2y}}dy$.



And I found the integral of both sides:



$int frac{1}{x}dx = int frac{e^{2y}}{1+e^{2y}}dy $



$ln(x)=frac{ln(e^{2y}+1)}{2}+c$.



I re-order it, and I still can't get any of the solutions... What I'm doing wrong?






calculus ordinary-differential-equations




calculus ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 23:07









parishiltonparishilton

31710




31710












  • $begingroup$
    What do the possible solutions look like?
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:29










  • $begingroup$
    You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:32




















  • $begingroup$
    What do the possible solutions look like?
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:29










  • $begingroup$
    You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
    $endgroup$
    – randomgirl
    Dec 20 '18 at 23:32


















$begingroup$
What do the possible solutions look like?
$endgroup$
– randomgirl
Dec 20 '18 at 23:29




$begingroup$
What do the possible solutions look like?
$endgroup$
– randomgirl
Dec 20 '18 at 23:29












$begingroup$
You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
$endgroup$
– randomgirl
Dec 20 '18 at 23:32






$begingroup$
You could write your $c$ differently like $ln(k)$ . You could then use some properties of log to rewrite your solution from there.
$endgroup$
– randomgirl
Dec 20 '18 at 23:32












1 Answer
1






active

oldest

votes


















1












$begingroup$

Just try to solve for $y$ using logarithm and exponential rules. Below, I use the notation $c$ for a constant $c in mathbb R$ not the same between each step (as well known, any form of a constant can be re-written as simply a constant) :



$$2ln(x) = ln(e^{2y}+1) + c Leftrightarrow e^{2y}+1 = e^{2ln(x)-c} Leftrightarrow e^{2y}+1 = cx^2$$
$$Leftrightarrow$$
$$e^{2y}=cx^2-1 implies y = frac{1}{2}ln(cx^2-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much!!! :-)
    $endgroup$
    – parishilton
    Dec 20 '18 at 23:58











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Just try to solve for $y$ using logarithm and exponential rules. Below, I use the notation $c$ for a constant $c in mathbb R$ not the same between each step (as well known, any form of a constant can be re-written as simply a constant) :



$$2ln(x) = ln(e^{2y}+1) + c Leftrightarrow e^{2y}+1 = e^{2ln(x)-c} Leftrightarrow e^{2y}+1 = cx^2$$
$$Leftrightarrow$$
$$e^{2y}=cx^2-1 implies y = frac{1}{2}ln(cx^2-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much!!! :-)
    $endgroup$
    – parishilton
    Dec 20 '18 at 23:58
















1












$begingroup$

Just try to solve for $y$ using logarithm and exponential rules. Below, I use the notation $c$ for a constant $c in mathbb R$ not the same between each step (as well known, any form of a constant can be re-written as simply a constant) :



$$2ln(x) = ln(e^{2y}+1) + c Leftrightarrow e^{2y}+1 = e^{2ln(x)-c} Leftrightarrow e^{2y}+1 = cx^2$$
$$Leftrightarrow$$
$$e^{2y}=cx^2-1 implies y = frac{1}{2}ln(cx^2-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much!!! :-)
    $endgroup$
    – parishilton
    Dec 20 '18 at 23:58














1












1








1





$begingroup$

Just try to solve for $y$ using logarithm and exponential rules. Below, I use the notation $c$ for a constant $c in mathbb R$ not the same between each step (as well known, any form of a constant can be re-written as simply a constant) :



$$2ln(x) = ln(e^{2y}+1) + c Leftrightarrow e^{2y}+1 = e^{2ln(x)-c} Leftrightarrow e^{2y}+1 = cx^2$$
$$Leftrightarrow$$
$$e^{2y}=cx^2-1 implies y = frac{1}{2}ln(cx^2-1)$$






share|cite|improve this answer









$endgroup$



Just try to solve for $y$ using logarithm and exponential rules. Below, I use the notation $c$ for a constant $c in mathbb R$ not the same between each step (as well known, any form of a constant can be re-written as simply a constant) :



$$2ln(x) = ln(e^{2y}+1) + c Leftrightarrow e^{2y}+1 = e^{2ln(x)-c} Leftrightarrow e^{2y}+1 = cx^2$$
$$Leftrightarrow$$
$$e^{2y}=cx^2-1 implies y = frac{1}{2}ln(cx^2-1)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 23:37









RebellosRebellos

15.5k31250




15.5k31250












  • $begingroup$
    Thank you so much!!! :-)
    $endgroup$
    – parishilton
    Dec 20 '18 at 23:58


















  • $begingroup$
    Thank you so much!!! :-)
    $endgroup$
    – parishilton
    Dec 20 '18 at 23:58
















$begingroup$
Thank you so much!!! :-)
$endgroup$
– parishilton
Dec 20 '18 at 23:58




$begingroup$
Thank you so much!!! :-)
$endgroup$
– parishilton
Dec 20 '18 at 23:58


















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