How to choose between two solutions where both seem to be correct?












2












$begingroup$


Question to solve: $ ysin(2x)dx - (y^2 + cos^2x)dy = 0 $ -----(i)



Writing (i) in exact form [$ M(x,y)dx + N(x,y)dy = 0$], we get



$ysin(2x)dx + [-(y^2 + cos^2x)]dy = 0$ $where,M=ysin(2x)$ and $N=[-(y^2 + cos^2x)]$




Since the given differential equation is exact, i.e, $∂M/∂y = ∂N/∂x = sin(2x)$



it's general solution is given by
$∫Mdx$ (with y constant) + $∫Ndy $ ("N" free from x terms) $=C$
My workout:



$∫Mdx$ (with y constant) = $∫ysin(2x)dx$ = $frac{-ycos(2x)}2$



$∫Ndy $ ("N" free from x terms)$ = ∫-y^2 dy = frac{-y^3}3 $



So, the general solution is $frac{-ycos(2x)}2$ + $frac{-y^3}3 $ $= C$



But if we differentiate the general solution it does not trace back to original Differential Equation.
However if I simplify $ cos^2(x)$ as $frac{cos2x+1}2$ I receive [$frac{-ycos(2x)}2$ + $frac{-y^3}3 - frac{y}2$ $= C$] as general solution and it does trace back to original differential eqaution. So, the problem is do i need verify everytime whether the obtained solution is correct or not, or both of the solutions are correct here?






calculus ordinary-differential-equations proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think checking the final solution is necessary, sometimes calculations have typos!
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:47










  • $begingroup$
    I believe there are no typos in either of the solution. But still your advice is useful.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 3:49
















2












$begingroup$


Question to solve: $ ysin(2x)dx - (y^2 + cos^2x)dy = 0 $ -----(i)



Writing (i) in exact form [$ M(x,y)dx + N(x,y)dy = 0$], we get



$ysin(2x)dx + [-(y^2 + cos^2x)]dy = 0$ $where,M=ysin(2x)$ and $N=[-(y^2 + cos^2x)]$




Since the given differential equation is exact, i.e, $∂M/∂y = ∂N/∂x = sin(2x)$



it's general solution is given by
$∫Mdx$ (with y constant) + $∫Ndy $ ("N" free from x terms) $=C$
My workout:



$∫Mdx$ (with y constant) = $∫ysin(2x)dx$ = $frac{-ycos(2x)}2$



$∫Ndy $ ("N" free from x terms)$ = ∫-y^2 dy = frac{-y^3}3 $



So, the general solution is $frac{-ycos(2x)}2$ + $frac{-y^3}3 $ $= C$



But if we differentiate the general solution it does not trace back to original Differential Equation.
However if I simplify $ cos^2(x)$ as $frac{cos2x+1}2$ I receive [$frac{-ycos(2x)}2$ + $frac{-y^3}3 - frac{y}2$ $= C$] as general solution and it does trace back to original differential eqaution. So, the problem is do i need verify everytime whether the obtained solution is correct or not, or both of the solutions are correct here?






calculus ordinary-differential-equations proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think checking the final solution is necessary, sometimes calculations have typos!
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:47










  • $begingroup$
    I believe there are no typos in either of the solution. But still your advice is useful.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 3:49














2












2








2





$begingroup$


Question to solve: $ ysin(2x)dx - (y^2 + cos^2x)dy = 0 $ -----(i)



Writing (i) in exact form [$ M(x,y)dx + N(x,y)dy = 0$], we get



$ysin(2x)dx + [-(y^2 + cos^2x)]dy = 0$ $where,M=ysin(2x)$ and $N=[-(y^2 + cos^2x)]$




Since the given differential equation is exact, i.e, $∂M/∂y = ∂N/∂x = sin(2x)$



it's general solution is given by
$∫Mdx$ (with y constant) + $∫Ndy $ ("N" free from x terms) $=C$
My workout:



$∫Mdx$ (with y constant) = $∫ysin(2x)dx$ = $frac{-ycos(2x)}2$



$∫Ndy $ ("N" free from x terms)$ = ∫-y^2 dy = frac{-y^3}3 $



So, the general solution is $frac{-ycos(2x)}2$ + $frac{-y^3}3 $ $= C$



But if we differentiate the general solution it does not trace back to original Differential Equation.
However if I simplify $ cos^2(x)$ as $frac{cos2x+1}2$ I receive [$frac{-ycos(2x)}2$ + $frac{-y^3}3 - frac{y}2$ $= C$] as general solution and it does trace back to original differential eqaution. So, the problem is do i need verify everytime whether the obtained solution is correct or not, or both of the solutions are correct here?






calculus ordinary-differential-equations proof-verification







share|cite|improve this question









$endgroup$




Question to solve: $ ysin(2x)dx - (y^2 + cos^2x)dy = 0 $ -----(i)



Writing (i) in exact form [$ M(x,y)dx + N(x,y)dy = 0$], we get



$ysin(2x)dx + [-(y^2 + cos^2x)]dy = 0$ $where,M=ysin(2x)$ and $N=[-(y^2 + cos^2x)]$




Since the given differential equation is exact, i.e, $∂M/∂y = ∂N/∂x = sin(2x)$



it's general solution is given by
$∫Mdx$ (with y constant) + $∫Ndy $ ("N" free from x terms) $=C$
My workout:



$∫Mdx$ (with y constant) = $∫ysin(2x)dx$ = $frac{-ycos(2x)}2$



$∫Ndy $ ("N" free from x terms)$ = ∫-y^2 dy = frac{-y^3}3 $



So, the general solution is $frac{-ycos(2x)}2$ + $frac{-y^3}3 $ $= C$



But if we differentiate the general solution it does not trace back to original Differential Equation.
However if I simplify $ cos^2(x)$ as $frac{cos2x+1}2$ I receive [$frac{-ycos(2x)}2$ + $frac{-y^3}3 - frac{y}2$ $= C$] as general solution and it does trace back to original differential eqaution. So, the problem is do i need verify everytime whether the obtained solution is correct or not, or both of the solutions are correct here?






calculus ordinary-differential-equations proof-verification




calculus ordinary-differential-equations proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 3:23









Abbas MiyaAbbas Miya

1338




1338












  • $begingroup$
    I think checking the final solution is necessary, sometimes calculations have typos!
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:47










  • $begingroup$
    I believe there are no typos in either of the solution. But still your advice is useful.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 3:49


















  • $begingroup$
    I think checking the final solution is necessary, sometimes calculations have typos!
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:47










  • $begingroup$
    I believe there are no typos in either of the solution. But still your advice is useful.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 3:49
















$begingroup$
I think checking the final solution is necessary, sometimes calculations have typos!
$endgroup$
– Nosrati
Dec 7 '18 at 3:47




$begingroup$
I think checking the final solution is necessary, sometimes calculations have typos!
$endgroup$
– Nosrati
Dec 7 '18 at 3:47












$begingroup$
I believe there are no typos in either of the solution. But still your advice is useful.
$endgroup$
– Abbas Miya
Dec 7 '18 at 3:49




$begingroup$
I believe there are no typos in either of the solution. But still your advice is useful.
$endgroup$
– Abbas Miya
Dec 7 '18 at 3:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):




$int Ndy$ ("N" free from x terms)$= ∫-y^2 dy = frac{-y^3}3$




A few lines earlier you stated (correctly!) that $N=[-(y^2+cos^2x)]$, therefore the integral of $N$ is
$$int N,dy=int[-(y^2+cos^2x)],dy,$$
not what you set up above, as the function you put inside the integral is not $N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 10:25













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%2f3029424%2fhow-to-choose-between-two-solutions-where-both-seem-to-be-correct%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









1












$begingroup$

The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):




$int Ndy$ ("N" free from x terms)$= ∫-y^2 dy = frac{-y^3}3$




A few lines earlier you stated (correctly!) that $N=[-(y^2+cos^2x)]$, therefore the integral of $N$ is
$$int N,dy=int[-(y^2+cos^2x)],dy,$$
not what you set up above, as the function you put inside the integral is not $N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 10:25


















1












$begingroup$

The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):




$int Ndy$ ("N" free from x terms)$= ∫-y^2 dy = frac{-y^3}3$




A few lines earlier you stated (correctly!) that $N=[-(y^2+cos^2x)]$, therefore the integral of $N$ is
$$int N,dy=int[-(y^2+cos^2x)],dy,$$
not what you set up above, as the function you put inside the integral is not $N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 10:25
















1












1








1





$begingroup$

The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):




$int Ndy$ ("N" free from x terms)$= ∫-y^2 dy = frac{-y^3}3$




A few lines earlier you stated (correctly!) that $N=[-(y^2+cos^2x)]$, therefore the integral of $N$ is
$$int N,dy=int[-(y^2+cos^2x)],dy,$$
not what you set up above, as the function you put inside the integral is not $N$.






share|cite|improve this answer









$endgroup$



The following line in your original solution is wrong, which is the reason the answer you obtained is wrong (and thus doesn't satisfy the original equation):




$int Ndy$ ("N" free from x terms)$= ∫-y^2 dy = frac{-y^3}3$




A few lines earlier you stated (correctly!) that $N=[-(y^2+cos^2x)]$, therefore the integral of $N$ is
$$int N,dy=int[-(y^2+cos^2x)],dy,$$
not what you set up above, as the function you put inside the integral is not $N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 4:06









zipirovichzipirovich

11.2k11631




11.2k11631












  • $begingroup$
    Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 10:25




















  • $begingroup$
    Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
    $endgroup$
    – Abbas Miya
    Dec 7 '18 at 10:25


















$begingroup$
Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
$endgroup$
– Abbas Miya
Dec 7 '18 at 10:25






$begingroup$
Is it not that while integrating "N", we should only integrate the terms in N which are free from x terms? And, N has $[−(y^2+cos^2x)]$, freeing it from x terms we are left with $-y^2$ only.
$endgroup$
– Abbas Miya
Dec 7 '18 at 10:25




















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%2f3029424%2fhow-to-choose-between-two-solutions-where-both-seem-to-be-correct%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more