Дифференциальные уравнения

Question 1

Question 2

Решите задачу:

$1)\; \; y'+y+7=0\\\\\frac{dy}{dx}=-(y+7)\\\\\int \frac{dy}{y+7}=-\int dx\\\\ln|y+7|=-x+C\\\\2)\; \; (\sqrt{xy}+\sqrt{x})dy=y\, dx\\\\\sqrt{x}(\sqrt{y}+1)dy=y\, dx\\\\\int \frac{\sqrt{y}+1}{y}\, dy= \int \frac{dx}{\sqrt{x}} \\\\\int (\frac{1}{\sqrt{y}}+\frac{1}{y})dy=\int \frac{dx}{\sqrt{x}}\\\\2\sqrt{y}+ln|y|=2\sqrt{x}+C\\\\y(0)=1\; \; \to \; \; 2+ln1=0+C\; ,\; C=2\\\\2\sqrt{y}+ln|y|=2\sqrt{x}+2$

$3)\; \; \frac{xy'-y}{x} =\frac{y}{x} \\\\y'-\frac{y}{x}=ctg\frac{y}{x}\\\\t=\frac{y}{x}\; ,\; \; y=tx\; ,\; \; y'=t'x+t\\\\t'x+t-t=ctgt\\\\\frac{dt}{dx}\cdot x=ctgt\\\\\frac{dt}{ctgt}=\frac{dx}{x}\\\\\int tgt\, dt=\int \frac{dx}{x}\\\\-ln|cost|=ln|x|+ln|C|\\\\\frac{1}{cost}=Cx\; \; \to \; \; cos\frac{y}{x}=\frac{1}{Cx}$

$4)\; \; xy'-x^2\cdot sinx=y\; |:x\ne 0\\\\y'-x\cdot sinx=\frac{y}{x}\\\\y'+\frac{y}{x}=x\cdot sinx\\\\y=uv\; ,\; \; y'=u'v+uv'\\\\u'v+uv'+\frac{uv}{x}=x\cdot sinx\\\\u'v+u(v'+\frac{v}{x})=x\cdot sinx\\\\a)\; \; v'+\frac{v}{x}=0,\; \; \frac{dv}{dx}=-\frac{v}{x}\\\\\frac{dv}{v}=-\frac{dx}{x}\; \; \; \to \; \; \; lnv=-lnx\; ,\; \; \; v=\frac{1}{x}\\\\b)\; \; u'\cdot \frac{1}{x}=x\cdot sinx\\\\\int du=\int x^2\cdot sinx\, dx\\\\u=-x^2\cdot cosx+2\int x\cdot cosx\, dx\\\\u=-x^2\cdot cosx+2(x\cdot sinx-\int sinx\, dx)$

$u=-x^2cosx+2x\cdot sinx-2cosx+C\\\\c)\; \; y=\frac{1}{x}(-x^2cosx+2x\cdot sinx-2cosx+C)\\\\y=-x\cdot cosx+2sinx-2\cdot \frac{cosx}{x}+\frac{C}{x}$