Solving the next differential equations












-2












$begingroup$


I'm having trouble resolving this equations. I know i have to substitute $y$ to $z$ so in the first one: $z' = y''$ ; $z = y'$ and it would be like: $z'+z'^2 = 1$. But i don't know what to do next





  • $yy''+y'^2=1$

  • $4y'' = xy'^{2}$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 5:27










  • $begingroup$
    @Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 5:32








  • 2




    $begingroup$
    For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:11


















-2












$begingroup$


I'm having trouble resolving this equations. I know i have to substitute $y$ to $z$ so in the first one: $z' = y''$ ; $z = y'$ and it would be like: $z'+z'^2 = 1$. But i don't know what to do next





  • $yy''+y'^2=1$

  • $4y'' = xy'^{2}$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 5:27










  • $begingroup$
    @Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 5:32








  • 2




    $begingroup$
    For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:11
















-2












-2








-2





$begingroup$


I'm having trouble resolving this equations. I know i have to substitute $y$ to $z$ so in the first one: $z' = y''$ ; $z = y'$ and it would be like: $z'+z'^2 = 1$. But i don't know what to do next





  • $yy''+y'^2=1$

  • $4y'' = xy'^{2}$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




I'm having trouble resolving this equations. I know i have to substitute $y$ to $z$ so in the first one: $z' = y''$ ; $z = y'$ and it would be like: $z'+z'^2 = 1$. But i don't know what to do next





  • $yy''+y'^2=1$

  • $4y'' = xy'^{2}$






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 Dec 6 '18 at 6:02







Neto_Lozano

















asked Dec 6 '18 at 5:05









Neto_LozanoNeto_Lozano

41




41












  • $begingroup$
    $z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 5:27










  • $begingroup$
    @Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 5:32








  • 2




    $begingroup$
    For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:11




















  • $begingroup$
    $z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 5:27










  • $begingroup$
    @Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 5:32








  • 2




    $begingroup$
    For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
    $endgroup$
    – Mattos
    Dec 6 '18 at 6:11


















$begingroup$
$z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
$endgroup$
– Nosrati
Dec 6 '18 at 5:27




$begingroup$
$z'^2+z'-1=0$ means $z'=dfrac{-1pmsqrt{5}}{2}$.
$endgroup$
– Nosrati
Dec 6 '18 at 5:27












$begingroup$
@Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
$endgroup$
– Neto_Lozano
Dec 6 '18 at 5:32






$begingroup$
@Nosrati I don't think is Cauchy-Euler. The Possible outcomes are: $y^2 = x^2+c2$ ; $y^2 = c1 + x^2$ ; $y^2 = x^2$ ; $y^2 - c1 = (x+c2)^2$
$endgroup$
– Neto_Lozano
Dec 6 '18 at 5:32






2




2




$begingroup$
For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
$endgroup$
– Mattos
Dec 6 '18 at 6:11






$begingroup$
For the first $$yy'' + y'^{2} = (yy')' = left( frac{1}{2} y^{2} right)''$$ For the second, let $y' = z$.
$endgroup$
– Mattos
Dec 6 '18 at 6:11












1 Answer
1






active

oldest

votes


















0












$begingroup$

Considering the first equation $$y''+y'^2=1$$ As you wrote, let $z=y'$ and now consider
$$z'+z^2=1$$ Now, inverse ($x$ being the variable) as
$$frac 1 {x'}+z^2=1implies x'=frac1{1-z^2}implies x+C=int frac{dz}{1-z^2}=frac 12log left(frac{1+z}{1-z}right)$$ Now, extract $z$ and ... continue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did a small modification, instead of $y''$ is $yy''$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 6:05










  • $begingroup$
    @Neto_Lozano. Make $y=sqrt z$ and look what happens.
    $endgroup$
    – Claude Leibovici
    Dec 6 '18 at 6:09











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Considering the first equation $$y''+y'^2=1$$ As you wrote, let $z=y'$ and now consider
$$z'+z^2=1$$ Now, inverse ($x$ being the variable) as
$$frac 1 {x'}+z^2=1implies x'=frac1{1-z^2}implies x+C=int frac{dz}{1-z^2}=frac 12log left(frac{1+z}{1-z}right)$$ Now, extract $z$ and ... continue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did a small modification, instead of $y''$ is $yy''$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 6:05










  • $begingroup$
    @Neto_Lozano. Make $y=sqrt z$ and look what happens.
    $endgroup$
    – Claude Leibovici
    Dec 6 '18 at 6:09
















0












$begingroup$

Considering the first equation $$y''+y'^2=1$$ As you wrote, let $z=y'$ and now consider
$$z'+z^2=1$$ Now, inverse ($x$ being the variable) as
$$frac 1 {x'}+z^2=1implies x'=frac1{1-z^2}implies x+C=int frac{dz}{1-z^2}=frac 12log left(frac{1+z}{1-z}right)$$ Now, extract $z$ and ... continue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did a small modification, instead of $y''$ is $yy''$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 6:05










  • $begingroup$
    @Neto_Lozano. Make $y=sqrt z$ and look what happens.
    $endgroup$
    – Claude Leibovici
    Dec 6 '18 at 6:09














0












0








0





$begingroup$

Considering the first equation $$y''+y'^2=1$$ As you wrote, let $z=y'$ and now consider
$$z'+z^2=1$$ Now, inverse ($x$ being the variable) as
$$frac 1 {x'}+z^2=1implies x'=frac1{1-z^2}implies x+C=int frac{dz}{1-z^2}=frac 12log left(frac{1+z}{1-z}right)$$ Now, extract $z$ and ... continue.






share|cite|improve this answer









$endgroup$



Considering the first equation $$y''+y'^2=1$$ As you wrote, let $z=y'$ and now consider
$$z'+z^2=1$$ Now, inverse ($x$ being the variable) as
$$frac 1 {x'}+z^2=1implies x'=frac1{1-z^2}implies x+C=int frac{dz}{1-z^2}=frac 12log left(frac{1+z}{1-z}right)$$ Now, extract $z$ and ... continue.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 6:03









Claude LeiboviciClaude Leibovici

120k1157132




120k1157132












  • $begingroup$
    I did a small modification, instead of $y''$ is $yy''$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 6:05










  • $begingroup$
    @Neto_Lozano. Make $y=sqrt z$ and look what happens.
    $endgroup$
    – Claude Leibovici
    Dec 6 '18 at 6:09


















  • $begingroup$
    I did a small modification, instead of $y''$ is $yy''$
    $endgroup$
    – Neto_Lozano
    Dec 6 '18 at 6:05










  • $begingroup$
    @Neto_Lozano. Make $y=sqrt z$ and look what happens.
    $endgroup$
    – Claude Leibovici
    Dec 6 '18 at 6:09
















$begingroup$
I did a small modification, instead of $y''$ is $yy''$
$endgroup$
– Neto_Lozano
Dec 6 '18 at 6:05




$begingroup$
I did a small modification, instead of $y''$ is $yy''$
$endgroup$
– Neto_Lozano
Dec 6 '18 at 6:05












$begingroup$
@Neto_Lozano. Make $y=sqrt z$ and look what happens.
$endgroup$
– Claude Leibovici
Dec 6 '18 at 6:09




$begingroup$
@Neto_Lozano. Make $y=sqrt z$ and look what happens.
$endgroup$
– Claude Leibovici
Dec 6 '18 at 6:09


















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