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






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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













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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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




















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