Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 388

\[1)\ y^{'} = \sin x;\text{\ \ \ y}(0) = 0:\]

\[y = - \cos x + C\]

\[0 = - \cos 0 + C\]

\[C = 1.\]

\[Ответ:\ \ y = - \cos x + 1.\]

\[2)\ y^{'} = 2\cos x;\text{\ \ \ y}(\pi) = 1:\]

\[y = 2\sin x + C\]

\[1 = 2\sin\pi + C\]

\[C = 1.\]

\[Ответ:\ \ y = 2\sin x + 1.\]

\[3)\ y^{'} = 3x^{2} + 4x - 1;y(1) = - 2:\]

\[y = 3 \bullet \frac{x^{3}}{3} + 4 \bullet \frac{x^{2}}{2} - x + C =\]

\[= x^{3} + 2x^{2} - x + C\]

\[- 2 = 1^{3} + 2 \bullet 1^{2} - 1 + C\]

\[C = - 4.\]

\[Ответ:\ \ y = x^{3} + 2x^{2} - x - 4.\]

\[4)\ y^{'} = 2 + 2x - 3x^{2};\ y( - 1) = 2:\]

\[y = 2x + 2 \bullet \frac{x^{2}}{2} - 3 \bullet \frac{x^{3}}{3} + C =\]

\[= 2x + x^{2} - x^{3} + C\]

\[2 = 2 \bullet ( - 1) + ( - 1)^{2} - ( - 1)^{3} + C\]

\[C = 2.\]

\[Ответ:\ \ y = 2x + x^{2} - x^{3} + 2.\]

\[5)\ y^{'} = e^{x};\text{\ \ \ y}(1) = 1:\]

\[y = e^{x} + C\]

\[1 = e^{1} + C\]

\[C = 1 - e.\]

\[Ответ:\ \ y = e^{x} + 1 - e.\]

\[6)\ y^{'} = e^{- x};\text{\ \ \ y}(0) = 2:\]

\[y = - e^{- x} + C\]

\[2 = - e^{0} + C\]

\[C = 3.\]

\[Ответ:\ \ y = - e^{- x} + 3.\]