Discrepancy in solutions of differential equation?
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1
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The differential equation at hand is this :
$$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
And what I have tried is this :
$$
int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$
and
$$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$
And the solution of this elementary problem comes out to be :
$$ psileft(xright)=Ae^{-2cosh^{-2}left(xright)}$$
But clearly,
$$ psileft(xright) = cosh^{-2}left(xright) $$
is a solution. But why can't I find it through integration?
differential-equations hyperbolic-functions
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up vote
1
down vote
favorite
The differential equation at hand is this :
$$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
And what I have tried is this :
$$
int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$
and
$$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$
And the solution of this elementary problem comes out to be :
$$ psileft(xright)=Ae^{-2cosh^{-2}left(xright)}$$
But clearly,
$$ psileft(xright) = cosh^{-2}left(xright) $$
is a solution. But why can't I find it through integration?
differential-equations hyperbolic-functions
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The differential equation at hand is this :
$$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
And what I have tried is this :
$$
int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$
and
$$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$
And the solution of this elementary problem comes out to be :
$$ psileft(xright)=Ae^{-2cosh^{-2}left(xright)}$$
But clearly,
$$ psileft(xright) = cosh^{-2}left(xright) $$
is a solution. But why can't I find it through integration?
differential-equations hyperbolic-functions
The differential equation at hand is this :
$$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
And what I have tried is this :
$$
int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$
and
$$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$
And the solution of this elementary problem comes out to be :
$$ psileft(xright)=Ae^{-2cosh^{-2}left(xright)}$$
But clearly,
$$ psileft(xright) = cosh^{-2}left(xright) $$
is a solution. But why can't I find it through integration?
differential-equations hyperbolic-functions
differential-equations hyperbolic-functions
edited Nov 14 at 11:19
Robert Z
90k1056128
90k1056128
asked Nov 14 at 11:00
Jalaj Chaturvedi
787
787
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1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Actually, we have that
$$inttanh(x) dx=intfrac{d(cosh(x))}{cosh(x)}=ln(cosh(x))+c$$
and
$$D(tanh(x))=frac{1}{cosh^2(x)}.$$
So you confused the derivative with the integral...
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Actually, we have that
$$inttanh(x) dx=intfrac{d(cosh(x))}{cosh(x)}=ln(cosh(x))+c$$
and
$$D(tanh(x))=frac{1}{cosh^2(x)}.$$
So you confused the derivative with the integral...
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
add a comment |
up vote
1
down vote
accepted
Actually, we have that
$$inttanh(x) dx=intfrac{d(cosh(x))}{cosh(x)}=ln(cosh(x))+c$$
and
$$D(tanh(x))=frac{1}{cosh^2(x)}.$$
So you confused the derivative with the integral...
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Actually, we have that
$$inttanh(x) dx=intfrac{d(cosh(x))}{cosh(x)}=ln(cosh(x))+c$$
and
$$D(tanh(x))=frac{1}{cosh^2(x)}.$$
So you confused the derivative with the integral...
Actually, we have that
$$inttanh(x) dx=intfrac{d(cosh(x))}{cosh(x)}=ln(cosh(x))+c$$
and
$$D(tanh(x))=frac{1}{cosh^2(x)}.$$
So you confused the derivative with the integral...
edited Nov 14 at 11:17
answered Nov 14 at 11:04
Robert Z
90k1056128
90k1056128
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
add a comment |
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
Oh! Thanks, I didn't thought about that.
– Jalaj Chaturvedi
Nov 14 at 11:12
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
@JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
– Robert Z
Nov 14 at 11:32
add a comment |
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