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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Inseparable differential equation ： ウィキペディア英語版 Inseparable differential equation In mathematics, an inseparable differential equation is an ordinary differential equation that cannot be solved by using separation of variables. To solve an inseparable differential equation one can employ a number of other methods, like the Laplace transform, substitution, etc. ==Examples== Consider the general inseparable equation :$\backslash frac+p(x)y=q(x)$ Now we will define a special factorial, ''μ'' as :$\backslash mu\; =\; e^$ Thus: :$\backslash frac=(e^)\backslash frac\; (\backslash int\; p(x)dx)$ :$\backslash frac=\backslash mu\; p(x)$ From here we can solve the equation using the above definition: :$\backslash mu\backslash frac+\backslash mu\; p(x)y\; =\; \backslash mu\; q(x)$ :$\backslash mu\backslash frac+y\backslash frac=\backslash mu\; q(x)$ (using the product rule in reverse) :$\backslash frac(\backslash mu\; y)\; =\; \backslash mu\; q(x)$ :$\backslash mu\; y\; =\; \backslash int\; \backslash mu\; q(x)dx$ Finally, we obtain: :$y=\backslash frac$ This can be used to solve most all inseparable equations containing no ''y'' to a degree other than one. For example, solving the inseparable equation: :$\backslash frac=x+y$ :$\backslash frac-y=x$ By arranging in the form required, we obtain: :$p(x)=-1\backslash$ :$q(x)=x\backslash$ :$\backslash frac+p(x)y=q(x)$ Now all that is necessary is to find the value of ''μ'' to plug into our original equation of $y=\backslash frac.$ :$\backslash mu\; =e^=e^=e^$ Plugging this into the original equation and simplifying gives us our final answer: :$y=\backslash frac\}$ :$y=e^x(-xe^-e^+C)\backslash$ :$y=Ce^-x-1\backslash$ Consider for example the inseparable equation :$2y\text{'}\text{'}+3y\text{'}+y=5.\backslash$ Let us solve it using the Laplace transform. One has that : $\backslash mathcal\backslash$ = s \mathcal\ - f(0) : $\backslash mathcal\backslash$ = s^2 \mathcal\ - s f(0) - f'(0) : $\backslash mathcal\backslash left\backslash$ = s^n \mathcal\ - s^ f(0) - \cdots - f^(0). Using the convenience that Laplace transforms follow the rules of linearity, one can solve the above example for ''y'' by performing a Laplace transform on both sides of the differential equation, substituting in the initial values, solving for the transformed function, and then performing an inverse transform. For the above example, assume initial values are $y(0)=0$ and $y\text{'}(0)=0.$ Then, : $2(s^2Y-s\backslash cdot\; 0-0)+3(s\; Y-0)+Y=\backslash frac.$ It follows that :$(2s+1)(s+1)Y=\backslash frac$ or :$Y=\backslash frac.$ Now one can just take the inverse Laplace transform of ''Y'' to get the solution ''y'' to the original equation. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Inseparable differential equation」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.