Question

Решить задачу Коши
y'+y/x=3x
y(1)=1
Срочно

Answer 1

$y'+y/x=3x \\ y=uv \\ y'=u'v+uv' \\\\ u'v+uv'+\frac{uv}{x}=3x \\ \left \{ {{v'+\frac{v}{x}=0} \atop {u'v=3x}} \right. \\ \frac{dv}{dx} =-\frac{v}{x} \\ \frac{dv}{v} =- \frac{dx}{x} \\ lnv=-lnx \\ v=\frac{1}{x} \\ \frac{du}{xdx}=3x \\ du=3x^2dx \\ u=x^3+C \\\\ y=uv=x^2+\frac{C}{x} \\\\ 1+\frac{C}{1}=1\\ C=0 \\\\ Y=x^2$

Answer 2

$\displaystyle y'+\frac{y}{x}=3x\\y=uv;y'=u'v+v'u\\u'v+v'u+\frac{1}{x}uv=3x\\u'v+u(v'+\frac{1}{x}v)=3x\\\begin{cases}v'+\frac{1}{x}v=0\\u'v=3x\end{cases}\\\frac{dv}{dx}+\frac{v}{x}=0|*\frac{dx}{v}\\\frac{dv}{v}=-\frac{dx}{x}\\\int\frac{dv}{v}=-\int\frac{dx}{x}\\ln|v|=-ln|x|\\v=\frac{1}{x}\\\frac{du}{xdx}=3x|*xdx\\du=3x^2dx\\\int du=3\int x^2dx\\u=x^3+C\\y=x^2+\frac{C}{x}\\y(1)=1\\1=1+C\\C=0\\\left[y=x^2\right]$