x = x^2...

July 2022 1 2 Report

диффернециальное уравнение.
1. найти частное решение : y = 3y / x , y(1)=1

2. Решить линейное уравнение: y` + y/x = x^2

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$1)\; \; y'=\frac{3y}{x}\; \; ,\; \; \; y(x=1)-1\\\\t= \frac{y}{x} \; ,\; \; y=tx\; ,\; y'=t'x+t\\\\t'x+t=3t\\\\t'x=2t\\\\\frac{dt}{dx}= \frac{2t}{x}\\\\\int \frac{dt}{t} =2\cdot \int \frac{dx}{x} \\\\ln|t|=2\cdot ln|x|+ln|C|\\\\t=Cx^2\\\\ \frac{y}{x}=Cx^2\\\\y=Cx^3$

$y(1)=1:\; \; 1=C\cdot 1^3\; \; \to \; \; \; C=1\\\\\underline {y=x^3}$

$2)\; \; y'+ \frac{y}{x}=x^2\\\\y=uv\; ,\; \; y'=u'v+uv'\\\\u'v+uv'+\frac{uv}{x}=x^2 \\\\u'v+u(v'+\frac{v}{x})=x^2\\\\1)\; \; v'+\frac{v}{x}=0\; \; ,\; \; \frac{dv}{dx}=-\frac{v}{x}\; \; ,\; \; \int \frac{dv}{v} =-\int \frac{dx}{x} \\\\ln|v|=-ln|x|\; ,\; \; v=x^{-1}=\frac{1}{x}\\\\2)\; \; u'\cdot \frac{1}{x}=x^2\\\\\frac{du}{dx}=x^3\; ,\; \; \int du=\int x^3\, dx\\\\u= \frac{x^4}{4}+C\\\\3)\; \; y= \frac{1}{x} \cdot ( \frac{x^4}{4}+C)\\\\y= \frac{x^3}{4}+\frac{C}{x}$

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