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A simple ordinary equation?
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How to solve the following ordinary equation?
$$-fracd^2f(x)dx^2+a f(x)^2=0,$$
where $a>0inmathbbR$.
Does anybody know of this equation?
differential-equations
asked Jul 31 at 14:52
Wein Eld
19710
add a comment |Â
up vote
0
down vote
favorite
How to solve the following ordinary equation?
$$-fracd^2f(x)dx^2+a f(x)^2=0,$$
where $a>0inmathbbR$.
Does anybody know of this equation?
differential-equations
asked Jul 31 at 14:52
Wein Eld
19710
Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to solve the following ordinary equation?
$$-fracd^2f(x)dx^2+a f(x)^2=0,$$
where $a>0inmathbbR$.
Does anybody know of this equation?
differential-equations
asked Jul 31 at 14:52
Wein Eld
19710
How to solve the following ordinary equation?
$$-fracd^2f(x)dx^2+a f(x)^2=0,$$
where $a>0inmathbbR$.
Does anybody know of this equation?
differential-equations
asked Jul 31 at 14:52
Wein Eld
19710
asked Jul 31 at 14:52
Wein Eld
19710
asked Jul 31 at 14:52
Wein Eld
19710
asked Jul 31 at 14:52
Wein Eld
19710
19710
Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11
add a comment |Â
Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11
Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11
add a comment |Â
1 Answer
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I will expand on LutzL's comment, and I will rewrite: $y=f(x)$:
$$-y''+ay^2=0.$$
Use the integrating factor $2y'$, multiplying it by every term to get:
$$-2y'y''+2ay'y^2=0.$$
Recognize that this is equivalent to:
$$fracddxleft((y')^2 + frac2a3y^3right)=0$$
by the chain rule. If a derivative is zero, then the function itself is constant, i.e.:
$$(y')^2+frac2a3y^3=c.$$
As LutzL mentioned, this is indeed separable; rearranging gives:
$$ fracdypmsqrtc-frac2a3y^3=dx $$
which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.
answered Jul 31 at 15:45
dbx
1,424111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I will expand on LutzL's comment, and I will rewrite: $y=f(x)$:
$$-y''+ay^2=0.$$
Use the integrating factor $2y'$, multiplying it by every term to get:
$$-2y'y''+2ay'y^2=0.$$
Recognize that this is equivalent to:
$$fracddxleft((y')^2 + frac2a3y^3right)=0$$
by the chain rule. If a derivative is zero, then the function itself is constant, i.e.:
$$(y')^2+frac2a3y^3=c.$$
As LutzL mentioned, this is indeed separable; rearranging gives:
$$ fracdypmsqrtc-frac2a3y^3=dx $$
which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.
answered Jul 31 at 15:45
dbx
1,424111
add a comment |Â
up vote
1
down vote
accepted
I will expand on LutzL's comment, and I will rewrite: $y=f(x)$:
$$-y''+ay^2=0.$$
Use the integrating factor $2y'$, multiplying it by every term to get:
$$-2y'y''+2ay'y^2=0.$$
Recognize that this is equivalent to:
$$fracddxleft((y')^2 + frac2a3y^3right)=0$$
by the chain rule. If a derivative is zero, then the function itself is constant, i.e.:
$$(y')^2+frac2a3y^3=c.$$
As LutzL mentioned, this is indeed separable; rearranging gives:
$$ fracdypmsqrtc-frac2a3y^3=dx $$
which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.
answered Jul 31 at 15:45
dbx
1,424111
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I will expand on LutzL's comment, and I will rewrite: $y=f(x)$:
$$-y''+ay^2=0.$$
Use the integrating factor $2y'$, multiplying it by every term to get:
$$-2y'y''+2ay'y^2=0.$$
Recognize that this is equivalent to:
$$fracddxleft((y')^2 + frac2a3y^3right)=0$$
by the chain rule. If a derivative is zero, then the function itself is constant, i.e.:
$$(y')^2+frac2a3y^3=c.$$
As LutzL mentioned, this is indeed separable; rearranging gives:
$$ fracdypmsqrtc-frac2a3y^3=dx $$
which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.
answered Jul 31 at 15:45
dbx
1,424111
I will expand on LutzL's comment, and I will rewrite: $y=f(x)$:
$$-y''+ay^2=0.$$
Use the integrating factor $2y'$, multiplying it by every term to get:
$$-2y'y''+2ay'y^2=0.$$
Recognize that this is equivalent to:
$$fracddxleft((y')^2 + frac2a3y^3right)=0$$
by the chain rule. If a derivative is zero, then the function itself is constant, i.e.:
$$(y')^2+frac2a3y^3=c.$$
As LutzL mentioned, this is indeed separable; rearranging gives:
$$ fracdypmsqrtc-frac2a3y^3=dx $$
which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.
answered Jul 31 at 15:45
dbx
1,424111
answered Jul 31 at 15:45
dbx
1,424111
answered Jul 31 at 15:45
dbx
1,424111
answered Jul 31 at 15:45
dbx
1,424111
1,424111
add a comment |Â
add a comment |Â
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Looks like this has something to do with the Weierstrass p function (wolframalpha.com/widgets/…). Where the solution is given by $f(x)=sqrt[3]frac6awpleft(sqrt[3]frac6a(x+c_1;0,c_2)right)$
– mrtaurho
Jul 31 at 15:00
As usual multiply with $2f'(x)$ and integrate to $-(f'(x))^2+frac2a3f(x)^3=C$. Then proceed with separation to reduce to a normal real integral.
– LutzL
Jul 31 at 15:11