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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028058%2fsolving-the-next-differential-equations%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028058%2fsolving-the-next-differential-equations%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more