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Solving $fracdydx=fracy^2x$
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The following differential equation is given:
$$fracdydx=fracy^2x$$
Separating variables and integrating:
$$int y^-2 dy = int x^-1 dx$$
$$-y^-1=ln|x|+c$$
$$y=frac1x$$
But the solution is given as:
$$y=frac1-c-ln x$$
How can this omission of the modulus be explained?
calculus integration differential-equations indefinite-integrals
asked Jul 19 at 16:20
FizzleDizzle
503
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up vote
1
down vote
favorite
The following differential equation is given:
$$fracdydx=fracy^2x$$
Separating variables and integrating:
$$int y^-2 dy = int x^-1 dx$$
$$-y^-1=ln|x|+c$$
$$y=frac1x$$
But the solution is given as:
$$y=frac1-c-ln x$$
How can this omission of the modulus be explained?
calculus integration differential-equations indefinite-integrals
asked Jul 19 at 16:20
FizzleDizzle
503
4
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following differential equation is given:
$$fracdydx=fracy^2x$$
Separating variables and integrating:
$$int y^-2 dy = int x^-1 dx$$
$$-y^-1=ln|x|+c$$
$$y=frac1x$$
But the solution is given as:
$$y=frac1-c-ln x$$
How can this omission of the modulus be explained?
calculus integration differential-equations indefinite-integrals
asked Jul 19 at 16:20
FizzleDizzle
503
The following differential equation is given:
$$fracdydx=fracy^2x$$
Separating variables and integrating:
$$int y^-2 dy = int x^-1 dx$$
$$-y^-1=ln|x|+c$$
$$y=frac1x$$
But the solution is given as:
$$y=frac1-c-ln x$$
How can this omission of the modulus be explained?
calculus integration differential-equations indefinite-integrals
asked Jul 19 at 16:20
FizzleDizzle
503
asked Jul 19 at 16:20
FizzleDizzle
503
asked Jul 19 at 16:20
FizzleDizzle
503
asked Jul 19 at 16:20
FizzleDizzle
503
503
4
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22
add a comment |Â
4
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22
4
4
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22
add a comment |Â
1 Answer
1
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oldest
votes
up vote
4
down vote
accepted
The solution given is for $x>0$ only. There is also the solution $$y=frac1-c-ln(-x)$$ for $x<0$ only. No solution may cross $x=0$.
answered Jul 19 at 16:22
vadim123
73.8k895184
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The solution given is for $x>0$ only. There is also the solution $$y=frac1-c-ln(-x)$$ for $x<0$ only. No solution may cross $x=0$.
answered Jul 19 at 16:22
vadim123
73.8k895184
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
add a comment |Â
up vote
4
down vote
accepted
The solution given is for $x>0$ only. There is also the solution $$y=frac1-c-ln(-x)$$ for $x<0$ only. No solution may cross $x=0$.
answered Jul 19 at 16:22
vadim123
73.8k895184
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The solution given is for $x>0$ only. There is also the solution $$y=frac1-c-ln(-x)$$ for $x<0$ only. No solution may cross $x=0$.
answered Jul 19 at 16:22
vadim123
73.8k895184
The solution given is for $x>0$ only. There is also the solution $$y=frac1-c-ln(-x)$$ for $x<0$ only. No solution may cross $x=0$.
answered Jul 19 at 16:22
vadim123
73.8k895184
answered Jul 19 at 16:22
vadim123
73.8k895184
answered Jul 19 at 16:22
vadim123
73.8k895184
answered Jul 19 at 16:22
vadim123
73.8k895184
73.8k895184
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
add a comment |Â
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
1
1
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
In particular, thanks to the discontinuity, the "c" in each half can be specified independently.
– IanF1
Jul 19 at 17:31
add a comment |Â
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4
The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,infty)$.
– Dylan
Jul 19 at 16:22