[tex](1 - {x}^{2} )y' + xy = 2x \\ ( 1 - {x}^{2} ) \frac{dy}{dx} = 2x - xy \\ \frac{1 - {x}^{2} }{dx} dy = - x(y - 2) \\ \frac{x}{(x - 1)(x + 1)} dx = \frac{1}{y - 2} dy \\ \int \frac{x}{(x - 1)(x + 1)} dx = \int \frac{1}{y - 2} dy \\ \frac{ln | {x}^{2} - 1 | }{2} = ln |y - 2| + ln |C| \\ {ln | {x}^{2} - 1 | }^{ \frac{1}{2} } = ln |C(y - 2)| \\ \sqrt{ {x}^{2} - 1 } = C(y - 2) \\ y - 2 = \frac{ \sqrt{ {x}^{2} - 1 } }{C} \\ y = \sqrt{ {x}^{2} - 1 } {C}^{ - 1} + 2 \\ y = 2 + C \sqrt{ {x}^{2} - 1} [/tex]
так как C → любое число, то выражение [tex] {C}^{-1} [/tex] и [tex] {C}^{1} [/tex] будут равны.
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[tex](1 - {x}^{2} )y' + xy = 2x \\ ( 1 - {x}^{2} ) \frac{dy}{dx} = 2x - xy \\ \frac{1 - {x}^{2} }{dx} dy = - x(y - 2) \\ \frac{x}{(x - 1)(x + 1)} dx = \frac{1}{y - 2} dy \\ \int \frac{x}{(x - 1)(x + 1)} dx = \int \frac{1}{y - 2} dy \\ \frac{ln | {x}^{2} - 1 | }{2} = ln |y - 2| + ln |C| \\ {ln | {x}^{2} - 1 | }^{ \frac{1}{2} } = ln |C(y - 2)| \\ \sqrt{ {x}^{2} - 1 } = C(y - 2) \\ y - 2 = \frac{ \sqrt{ {x}^{2} - 1 } }{C} \\ y = \sqrt{ {x}^{2} - 1 } {C}^{ - 1} + 2 \\ y = 2 + C \sqrt{ {x}^{2} - 1} [/tex]
так как C → любое число, то выражение [tex] {C}^{-1} [/tex] и [tex] {C}^{1} [/tex] будут равны.