Question

Дифференциальные уравнения

Answer 1

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

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

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

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