While solving LDE can the IF $e^{int frac{dx}{x}}$ be taken as x instead of |x|?
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While solving the differential equation
$xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
calculus ordinary-differential-equations
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
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add a comment |
$begingroup$
While solving the differential equation
$xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
calculus ordinary-differential-equations
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
$endgroup$
add a comment |
$begingroup$
While solving the differential equation
$xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
calculus ordinary-differential-equations
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
$endgroup$
While solving the differential equation
$xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
asked Jan 18 at 17:14
BibThePhysicistBibThePhysicist
397
397
add a comment |
add a comment |
1 Answer
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No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
$$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
that is
$$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
As you can see the value of the constant $Cnot =0$ is irrelevant.
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
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$begingroup$
No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
$$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
that is
$$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
As you can see the value of the constant $Cnot =0$ is irrelevant.
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
$endgroup$
add a comment |
$begingroup$
No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
$$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
that is
$$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
As you can see the value of the constant $Cnot =0$ is irrelevant.
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
$endgroup$
add a comment |
$begingroup$
No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
$$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
that is
$$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
As you can see the value of the constant $Cnot =0$ is irrelevant.
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
$endgroup$
No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
$$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
that is
$$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
As you can see the value of the constant $Cnot =0$ is irrelevant.
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
edited Jan 18 at 17:37
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
answered Jan 18 at 17:23
Robert ZRobert Z
102k1072145
102k1072145
add a comment |
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