積分因子メモ

$\displaystyle xy+\frac{y}{\sqrt{y^ 2-x^ 2}}+\left(y^ 2-\frac{x}{\sqrt{y^ 2-x^ 2}}\right)y^ {\prime}=0.$

【解答】

$P(x,\ y)=xy+y/\sqrt{y^ 2-x^ 2},$

$Q(x,\ y)=y^ 2-x/\sqrt{y^ 2-x^ 2}$

とおくと,

${ \displaystyle \begin{eqnarray} \frac{\partial P}{\partial y}&=&x+\frac{\sqrt{y^ 2-x^ 2}-y^ 2/\sqrt{y^ 2-x^ 2}}{y^ 2-x^ 2}\\ &=&x-\frac{x^ 2}{(y^ 2-x^ 2)^ {3/2},} \end{eqnarray} }$

${ \displaystyle \begin{eqnarray} \frac{\partial Q}{\partial x}&=&-\frac{\sqrt{y^ 2-x^ 2}-x(-x)/\sqrt{y^ 2-x^ 2}}{y^ 2-x^ 2}\\ &=&-\frac{y^ 2}{(y^ 2-x^ 2)^ {3/2}.} \end{eqnarray} }$

${ \displaystyle \begin{eqnarray} \therefore \frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}&=&x-\frac{x^ 2}{(y^ 2-x^ 2)^ {3/2}}+\frac{y^ 2}{(y^ 2-x^ 2)^ {3/2}}\\ &=&x+\frac{1}{\sqrt{y^ 2-x^ 2}} (\neq 0). \end{eqnarray} }$

よって全微分型ではないが, $(-1/P)(\partial P/\partial y-\partial Q/\partial x)=-1/y$ となるから積分因子として $\displaystyle \mu(y)=\exp\left[{\int{-\frac{dy}{y}}}\right]=1/y$ がとれる. すなわち,

$\displaystyle x+\frac{1}{\sqrt{y^ 2-x^ 2}}+\left(y-\frac{x}{y\sqrt{y^ 2-x^ 2}}\right)y^ {\prime}=0$

は全微分型.

${ \displaystyle \begin{eqnarray} & &\int _ {x _ 0} ^ {x}{\left(x+\frac{1}{\sqrt{y^ 2-x^ 2}}\right)dx}+\int _ {y _ 0}^ {y}{\left(y-\frac{x _ 0}{y\sqrt{y^ 2-{x _ 0}^ 2}}\right)dy}\\ &=&\left[\frac{x^ 2}{2}+\mathrm{Arcsin}\frac{x}{y}\right] _ {x _ 0} ^ {x}+\left[\frac{y^ 2}{2}\right] _ {y _ 0} ^ {y}-{x _ 0}\underline{\int _ {y _ 0} ^ {y}{\frac{dy}{y\sqrt{y^ 2 - {x _ 0}^ 2}}}} \ \ (\ast) \end{eqnarray} }$

下線部を計算するために $\displaystyle I(y)=\int{\frac{dy}{y\sqrt{y^ 2-{x _ 0}^ 2}}}$ を求める:

$t=y+\sqrt{y^ 2-{x _ 0}^ 2}$ とおくと,

$\displaystyle t^ 2-2yt+y^ 2=y^ 2-{x _ 0}^2.\ \therefore y=\frac{1}{2}\left(t+\frac{{x _ 0}^ 2}{t}\right).$

$\displaystyle \therefore \sqrt{y^ 2-{x _ 0}^ 2}=\frac{1}{2}\left(t-\frac{{x _ 0}^ 2}{t}\right),\ dy=\frac{1}{2}\left(1-\frac{{x _ 0}^ 2}{t^ 2}\right)dt.$

${ \displaystyle \begin{eqnarray} \therefore I(y) &=& \int{\frac{(1/2)(1-{x _ 0}^ 2/t^ 2)}{(1/4)(t+{x _ 0}^ 2/t)(t-{x _ 0}^ 2/t)}dt} \\ &=&\int{2\frac{1/t}{t+{x _ 0}^ 2/t}dt}=2\int{\frac{dt}{t^ 2+{x _ 0}^ 2}}\\ &=&(2/x _ 0)\mathrm{Arctan}(t/x _ 0)\\ &=&\frac{2}{x _ 0}\mathrm{Arctan}\frac{y+\sqrt{y^ 2-{x _ 0}^ 2}}{x _ 0}. \end{eqnarray} }$

ゆえに,

$\displaystyle (\ast)=\frac{x^ 2+y^ 2}{2}+\mathrm{Arcsin}\frac{x}{y}-\mathrm{Arcsin}\frac{x _ 0}{y}-2\mathrm{Arctan}\frac{y+\sqrt{y^ 2-{x _ 0}^ 2}}{x _ 0}+c _ 1$