First-Order ODE (2) - Exact ODE

4. Exact ODE

 

Total Differential

$ u(x,y)=c $

$ du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=u_{x}dx+u_{y}dy=0 $

 

Definition

$ M(x,y)dx+N(x,y)dy=0 $ is called an Exact ODE

$ \exists u(x,y) $ s.t. $ u_{x}=M, u_{y}=N $

 

Theorem

$ M(x,y)dx+N(x,y)dy=0 $ is exact $ \Leftrightarrow M_{y}=N_{x} $

 

$ u=\int M dx+k(y)=\int N dy+l(x) $

 

ex) $ (e^{x}+y)dx+(x-e^{-y})dy=0 $

$ M=e^{x}+y, N=x-e^{-y} $

$ M_{y}=1=N_{x} $: exact

$ u_{x}=M, u_{y}=N $

$ u=\int M dx=e^{x}+xy+k(y) $

$ u_{y}=x+k'(y)=x-e^{-y} $

$ k'(y)=-e^{-y} \rightarrow k(y)=e^{-y}+c_{1} $

$ u=e^{x}+xy+e^{-y}+c_{1}=c_{2} $

$ \therefore e^{x}+xy+e^{-y}=c $

 

 

5. Reduction to Exact Form (Intergrating Factor)

 

$ P(x,y)dx+Q(x,y)dy=0 $: Nonexact

$ FPdx+FQdy=0 $: Exact ($ F(x,y) $: Intergrating Factor)

 

$ (FP)_{y}=(FQ)_{x} $

$ F_{y}P+FP_{y}=F_{x}Q+FQ_{x} $

 

i) $ F=F(x) $

$ FP_{y}=F'Q+FQ_{x} $

$ \frac{1}{F}{\mathrm{d} F}=\frac{1}{Q}(P_{y}-Q_{x}){\mathrm{d} x} $

$ F=e^{\int R(x)dx} $ ($ R(x)=\frac{1}{Q}(P_{y}-Q_{x}) $)

 

ii) $F^{*}=F^{*}(y) $

$ F'^{*}P+FP_{y}=FQ_{x} $

$ \frac{1}{F^{*}}{\mathrm{d} F^{*}}=\frac{1}{P}(Q_{x}-P_{y}){\mathrm{d} x} $

$ F^{*}=e^{\int R^{*}(y)dy} $ ($ R^{*}(y)=\frac{1}{P}(Q_{x}-P_{y}) $)

 

ex) $ (e^{x+y}+ye^{y})dx+(xe^{y}-1)dy=0 $

$ P=e^{x+y}+ye^{y}, Q=xe^{y}-1 $

$ P_{y}=e^{x+y}+e^{y}+ye^{y}, Q_{x}=e^{y} $

$ P_{y}\neq Q_{x} $: not exact

$ R(x)=\frac{1}{Q}(P_{y}-Q_{x})=\frac{1}{xe^{y}-1}(e^{x+y}+e^{y}+ye^{y}-e^{y})=\frac{e^{x+y}+ye^{y}}{xe^{y}-1} $: y가 포함되어 있음

$ R^{*}(y)=\frac{1}{P}(Q_{x}-P_{y})=\frac{1}{e^{x+y}+ye^{y}}(e^{y}-e^{x+y}-e^{y}-ye^{y})=\frac{-(e^{x+y}+ye^{y})}{e^{x+y}+ye^{y}}=-1 $

$ F^{*}=e^{\int R^{*}(y)dy}=e^{\int -1dy}=e^{-y} $: intergrating factor

$ \therefore (e^{x}+y)dx+(x-e^{-y})dy=0 $: exact