upd

notes/notes.typ

@@ -297,4 +297,70 @@ $
 e^x + y dot x + sin y dot x + e^y eq C_2
 $
 
+=== Интегрирующий множитель
 
+$
+mu(x, y) - "интегрирующий множитель"
+$
+
+$
+mu(x, y) dot P(x, y) d x + mu(x, y) dot Q(x, y) d y eq 0
+$
+
+$
+frac(partial(mu(x, y) dot P(x, y)), partial y) eq frac(partial(mu(x, y) dot Q(x, y)), partial x)
+$
+
+$
+mu eq mu(x, y), space.quad P eq P(x, y), space.quad Q eq Q(x, y)
+$
+
+$
+frac(partial M, partial y) dot P + mu dot frac(partial P, partial y) eq frac(partial mu, partial x) dot Q + mu dot frac(partial Q, partial x)
+$
+
+$
+frac(partial mu, partial y) dot P - frac(partial mu, partial x) dot Q eq mu dot (frac(partial Q, partial x) - frac(partial P, partial y))
+$
+
+1) Пусть $mu eq mu(x)$
+
+$
+- frac(d mu, d x) dot Q eq mu dot (frac(partial Q, partial x) - frac(partial P, partial y))
+$
+
+$
+frac(d mu, mu) eq -1/Q (frac(partial Q, partial x) - frac(partial P, partial y)) dot d x
+$
+
+$
+ln mu eq integral 1/Q (frac(partial P, partial y) - frac(partial Q, partial x)) dot d x
+$
+
+$
+mu (x) eq exp(integral underbrace(1/Q (frac(partial P, partial y) - frac(partial Q, partial x)), eq F_1 (x)) dot d x)
+$
+
+При этом
+
+$
+1/Q (frac(partial P, partial y) - frac(partial Q, partial x)) eq F_1 (x)
+$
+
+То есть зависеть только от $x$.
+
+2) Пусть $mu eq mu(y)$
+
+Размышляя аналогично, получим формулу
+
+$
+mu(y) eq exp(integral underbrace(1/P (frac(partial Q, partial x) - frac(partial P, partial y)), eq F_2 (y)) d y)
+$
+
+При этом 
+
+$
+1/P (frac(partial Q, partial x) - frac(partial P, partial y)) eq F_2 (y)
+$
+
+То есть зависеть только от $y$.

подготовка к кр/main.pdf

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