Задача Коши [tex]y'+xy=- x^{3} [/tex] [tex]y(0)=3[/tex]. Created by GinV. matematika-ru

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July 2022 1 13 Report

Задача Коши
[tex]y'+xy=- x^{3} [/tex]
[tex]y(0)=3[/tex]

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NNNLLL54

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$y'+xy=-x^3\; ,\; \; y(0)=3\\\\y=uv\\\\u'v+uv'+xuv=-x^3\\\\u'v+u(v'+xv)=-x^3\\\\a)\; \; v'+xv=0\; \; ,\; \; \frac{dv}{dx}=-xv\; ,\; \; \int \frac{dv}{v}=-\int x\, dx\\\\ln|v|=-\frac{x^2}{2} \; \; \to \; \; \; v=e^{-\frac{x^2}{2}}\\\\b)\; \; u'\cdot e^{-\frac{x^2}{2}}=-x^3 \\\\\int du=-\int x^3\cdot e^{\frac{x^2}{2}}\, dx$

$\int x^3\cdot e^{\frac{x^2}{2}}\, dx=[\; u=x^2\; ,\; du=2x\, dx\; ,\; dv=x\cdot e^{\frac{x^2}{2}}\, dx\; ,\\\\v=\int x\cdot e^{\frac{x^2}{2}}\, dx=[\; t= \frac{x^2}{2} \; ,\; dt=\frac{1}{2}\cdot 2x\, dx=x\, dx\; ]=\\\\=\int e^{t}\, dt=e^{t}=e^{\frac{x^2}{2}}\; ]=$

$=x^2\cdot e^{\frac{x^2}{2}}-2\int x\cdot e^{ \frac{x^2}{2} }\, dx=x^2\cdot e^{\frac{x^2}{2}}-2e^{ \frac{x^2}{2} }+C_1\\\\u=-x^2e^ {\frac{x^2}{2}}+2e^ {\frac{x^2}{2} }+C\\\\c)\; \; y=e^{-\frac{x^2}{2}}\cdot (-x^2e^{\frac{x^2}{2}}+2e^{\frac{x^2}{2}}+C)=-x^2+2+Ce^{-\frac{x^2}{2}}\\\\d)\; \; y(0)=3\; ,\; \; 3=2+C\cdot e^0=2+C\; ,\; \; C=1\\\\y=-x^2+2+e^{-\frac{x^2}{2}}$

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