Mixing
Definite integration of an exact differential
$begingroup$
The following definitions are given:
$$du(hat{x},hat{y})=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{1}$$
$$0=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{2}$$
The task is to prove the equation:
$$u(x,y)=int_0^1[M(tx,ty)x+N(tx,ty)y]dt+Ctag{3}$$
where C is a constant.
My approach has been to start with (1) and to make the substitutions $hat{x}=xt$ and $hat{y}=yt$. As a result, $dhat{x}=xdt+tdx$, and an equivalent statement can be made for $dhat{y}$.
Integrating the LHS of Eq 1 over the interval $t=[0,1]$, I arrive at
$$int_{t=0}^{t=1}du(hat{x},hat{y})=int_{t=0}^{t=1}du(xt,yt)$$
$$=int_{u(0,0)}^{u(x,y)}du$$$$=u(x,y)-C_{LHS}$$
where $C_{LHS}=u(0,0)$. Rearranging and integrating Eq 1's RHS over the interval $t=[0,1]$,
$$int_{t=0}^{t=1}[M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}]=$$
$$=int_{t=0}^{t=1}[M(tx,ty)x+N(tx,ty)y]dt+
underbrace {int_{t=0}^{t=1}t[M(tx,ty)dx+N(tx,ty)dy]}$$
So all that's separating me from the desired answer is a proof that the underbraced bit is either zero or constant. As the inclusion of $t$ just scales equation 2 in a seemingly inconsequential way, I can see that being the case, but I can't seem to formally show this. Any insight? And how does my math look so far?
ordinary-differential-equations definite-integrals
asked Jan 26 at 20:40
ArnoldArnold
12
$endgroup$
add a comment |
$begingroup$
The following definitions are given:
$$du(hat{x},hat{y})=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{1}$$
$$0=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{2}$$
The task is to prove the equation:
$$u(x,y)=int_0^1[M(tx,ty)x+N(tx,ty)y]dt+Ctag{3}$$
where C is a constant.
My approach has been to start with (1) and to make the substitutions $hat{x}=xt$ and $hat{y}=yt$. As a result, $dhat{x}=xdt+tdx$, and an equivalent statement can be made for $dhat{y}$.
Integrating the LHS of Eq 1 over the interval $t=[0,1]$, I arrive at
$$int_{t=0}^{t=1}du(hat{x},hat{y})=int_{t=0}^{t=1}du(xt,yt)$$
$$=int_{u(0,0)}^{u(x,y)}du$$$$=u(x,y)-C_{LHS}$$
where $C_{LHS}=u(0,0)$. Rearranging and integrating Eq 1's RHS over the interval $t=[0,1]$,
$$int_{t=0}^{t=1}[M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}]=$$
$$=int_{t=0}^{t=1}[M(tx,ty)x+N(tx,ty)y]dt+
underbrace {int_{t=0}^{t=1}t[M(tx,ty)dx+N(tx,ty)dy]}$$
So all that's separating me from the desired answer is a proof that the underbraced bit is either zero or constant. As the inclusion of $t$ just scales equation 2 in a seemingly inconsequential way, I can see that being the case, but I can't seem to formally show this. Any insight? And how does my math look so far?
ordinary-differential-equations definite-integrals
asked Jan 26 at 20:40
ArnoldArnold
12
$endgroup$
add a comment |
$begingroup$
The following definitions are given:
$$du(hat{x},hat{y})=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{1}$$
$$0=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{2}$$
The task is to prove the equation:
$$u(x,y)=int_0^1[M(tx,ty)x+N(tx,ty)y]dt+Ctag{3}$$
where C is a constant.
My approach has been to start with (1) and to make the substitutions $hat{x}=xt$ and $hat{y}=yt$. As a result, $dhat{x}=xdt+tdx$, and an equivalent statement can be made for $dhat{y}$.
Integrating the LHS of Eq 1 over the interval $t=[0,1]$, I arrive at
$$int_{t=0}^{t=1}du(hat{x},hat{y})=int_{t=0}^{t=1}du(xt,yt)$$
$$=int_{u(0,0)}^{u(x,y)}du$$$$=u(x,y)-C_{LHS}$$
where $C_{LHS}=u(0,0)$. Rearranging and integrating Eq 1's RHS over the interval $t=[0,1]$,
$$int_{t=0}^{t=1}[M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}]=$$
$$=int_{t=0}^{t=1}[M(tx,ty)x+N(tx,ty)y]dt+
underbrace {int_{t=0}^{t=1}t[M(tx,ty)dx+N(tx,ty)dy]}$$
So all that's separating me from the desired answer is a proof that the underbraced bit is either zero or constant. As the inclusion of $t$ just scales equation 2 in a seemingly inconsequential way, I can see that being the case, but I can't seem to formally show this. Any insight? And how does my math look so far?
ordinary-differential-equations definite-integrals
asked Jan 26 at 20:40
ArnoldArnold
12
$endgroup$
The following definitions are given:
$$du(hat{x},hat{y})=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{1}$$
$$0=M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}tag{2}$$
The task is to prove the equation:
$$u(x,y)=int_0^1[M(tx,ty)x+N(tx,ty)y]dt+Ctag{3}$$
where C is a constant.
My approach has been to start with (1) and to make the substitutions $hat{x}=xt$ and $hat{y}=yt$. As a result, $dhat{x}=xdt+tdx$, and an equivalent statement can be made for $dhat{y}$.
Integrating the LHS of Eq 1 over the interval $t=[0,1]$, I arrive at
$$int_{t=0}^{t=1}du(hat{x},hat{y})=int_{t=0}^{t=1}du(xt,yt)$$
$$=int_{u(0,0)}^{u(x,y)}du$$$$=u(x,y)-C_{LHS}$$
where $C_{LHS}=u(0,0)$. Rearranging and integrating Eq 1's RHS over the interval $t=[0,1]$,
$$int_{t=0}^{t=1}[M(hat{x},hat{y})dhat{x}+N(hat{x},hat{y})dhat{y}]=$$
$$=int_{t=0}^{t=1}[M(tx,ty)x+N(tx,ty)y]dt+
underbrace {int_{t=0}^{t=1}t[M(tx,ty)dx+N(tx,ty)dy]}$$
So all that's separating me from the desired answer is a proof that the underbraced bit is either zero or constant. As the inclusion of $t$ just scales equation 2 in a seemingly inconsequential way, I can see that being the case, but I can't seem to formally show this. Any insight? And how does my math look so far?
ordinary-differential-equations definite-integrals
ordinary-differential-equations definite-integrals
asked Jan 26 at 20:40
ArnoldArnold
12
asked Jan 26 at 20:40
ArnoldArnold
12
edited Jan 26 at 20:51
Arnold
asked Jan 26 at 20:40
ArnoldArnold
12
asked Jan 26 at 20:40
ArnoldArnold
12
asked Jan 26 at 20:40
ArnoldArnold
12
12
add a comment |
add a comment |
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