In January 2025
Answer to Question #261820 in Differential Equations for blossomqt
π¦π ππ(π₯) β π₯π¦2 = π
differentiate both parts of the given equation respect to x:
"(\ud835\udc66\ud835\udc60\ud835\udc56\ud835\udc5b(\ud835\udc65) \u2212 \ud835\udc65\ud835\udc66^2)_x' =( \ud835\udc50)_x'"
we can get Differential Equation:
"y'sinx+ycosx-y^2-2yy'x=0"
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