المحلولة بالنسبة للتابع y

الكاتب Muhanad

المعادلات التفاضلية^[م] العادية من المرتبة الأولى والمحلولة بالنسبة للتابع من الشكل: $y= f(y')$

لحل هذه المعادلة نفرض أنّ $y' = t\quad$

$\begin{array}{l}\\\left. \begin{array}{l}y = f(y')\\\\y' = t\\\end{array} \right\}\quadÂ \Rightarrow \quad y = f(t)\quadÂ \Rightarrow \quad \left\{ \begin{array}{l}dy = f'(t)dt\\\\dy = y'dx = t\;dx\\\end{array} \right\}\\\\\\\Rightarrow \quad t\;dx = f'(t)dt\quadÂ \Rightarrow \quad x = \int {\frac{{f'(t)}}{t}\;dt}\\\end{array}$

وبالتالي الحل العام يعطى بالعلاقة :

$\boxed{\;\;\;\;\;\begin{array}{l}x =\int {\frac{{f'(t)}}{t}.dt + c}\\ \\y = f(t)\\\end{array} \;\;\;\;\;}$

نبذة عن كاتب الموضوع

$User picture$

Muhanad مشرف في الرياضيات رمز

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