Question

Найти общие и частные решения линейных дифференциальных уравнений первого порядка (х=uv)

Answer 1

Интегрирующий множитель:
$\it\displaystyle (xcos^2y-y^2)y'=ycos^2y\\ycos^2ydx-(xcos^2y-y^2)dy=0\\\frac{dP}{dy}=cos^2y-2ycosysiny\\\frac{dQ}{dx}=-cos^2y\\\frac{d\mu}{\mu}=-\frac{1}{ycos^2y}(cos^2y-2ycosysiny+cos^2y)dy\\\int\frac{d\mu}{\mu}=\int(-\frac{2}{y}+2tgy)dy\\ln|\mu|=-2ln|y|-2ln|cosy|\\\mu=\frac{1}{y^2cos^2y}\\\frac{dx}{y}-(\frac{x}{y^2}-\frac{1}{cos^2y})dy=0\\\frac{dP}{dy}=-\frac{1}{y^2}\\\frac{dQ}{dx}=-\frac{1}{y^2}$
$\it\displaystyle\begin{cases}\frac{dF}{dx}=\frac{1}{y}\\\frac{dF}{dy}=-(\frac{x}{y^2}-\frac{1}{cos^2y})\end{cases}\\F=\int\frac{1}{y}dx=\frac{x}{y}+\phi(y)\\\frac{dF}{dy}=-\frac{x}{y^2}+\phi'(y)=-\frac{x}{y^2}+\frac{1}{cos^2y}\\\phi'(y)=\frac{1}{cos^2y}\\\phi(y)=\int\frac{dy}{cos^2y}=tgy+C\\\frac{x}{y}+tgy+C=0\\y(\pi)=\frac{\pi}{4}\\\\4+1+C=0\\C=-5\\\\OTBET:\frac{x}{y}+tgy-5=0$
----------
$\it\displaystyle e^{y^2}(dx-2xydy)=ydy\\e^{y^2}dx-2xye^{y^2}dy=ydy\\e^{y^2}dx-(2xye^{y^2}+y)dy=0\\\frac{dP}{dy}=2ye^{y^2}\\\frac{dQ}{dx}=-2ye^{y^2}\\\frac{d\mu}{\mu}=-\frac{1}{e^{y^2}}(2ye^{y^2}+2ye^{y^2})dy\\\int\frac{d\mu}{\mu}=-\int4ydy\\ln|\mu|=-2y^2\\\mu=\frac{1}{e^{2y^2}}\\\\\frac{dx}{e^{y^2}}-(\frac{2xy}{e^{y^2}}+\frac{y}{e^{2y^2}})dy=0\\\frac{dP}{dy}=-\frac{2y}{e^{y^2}}\\\frac{dQ}{dx}=-\frac{2y}{e^{y^2}}$
$\it\displaystyle \begin{cases}\frac{dF}{dx}=\frac{1}{e^{y^2}}\\\frac{dF}{dy}=-(\frac{2xy}{e^{y^2}}+\frac{y}{e^{2y^2}})\end{cases}\\F=\int\frac{1}{e^{y^2}}dx=\frac{x}{e^{y^2}}+\phi(y)\\\frac{dF}{dy}=-x*e^{-y^2}*2y+\phi'(y)=-\frac{2xy}{e^{y^2}}+\phi'(y)\\-\frac{2xy}{e^{y^2}}+\phi'(y)=-(\frac{2xy}{e^{y^2}}+\frac{y}{e^{2y^2}})\\\phi'(y)=-\frac{y}{e^{y^2}}\\\phi(y)=-\int\frac{ydy}{e^{2y^2}}\\\phi(y)=\frac{1}{4}\int\frac{d(-2y^2)}{e^{2y^2}}=\frac{1}{4e^{2y^2}}+C\\\frac{x}{e^{y^2}}+\frac{1}{4e^{2y^2}}+C=0$
$y(0)=0\\\frac{0}{1}+\frac{1}{4}+C=0\\C=-\frac{1}{4}\\OTBET: \frac{x}{e^{y^2}}+\frac{1}{4e^{2y^2}}-\frac{1}{4}=0$
----------
$\it\displaystyle (104y^3-x)y'=4y\\4ydx-(104y^3-x)dy=0\\\frac{dP}{dy}=4\\\frac{dQ}{dx}=1\\\frac{d\mu}{\mu}=-\frac{1}{4y}(4-1)dy\\\int\frac{d\mu}{\mu}=-\frac{3}{4}\int\frac{dy}{y}\\ln|\mu|=-\frac{3}{4}ln|y|\\\mu=\frac{1}{y^\frac{3}{4}}\\4y^\frac{1}{4}-(104y^\frac{9}{4}-\frac{x}{y^\frac{3}{4}})=0\\\frac{dP}{dy}=\frac{1}{y^\frac{3}{4}}\\\frac{dQ}{dx}=\frac{1}{y^\frac{3}{4}}$
$\it\displaystyle \begin{cases}\frac{dF}{dx}=4y^\frac{1}{4}\\\frac{dF}{dy}=-(104y^\frac{9}{4}-\frac{x}{y^\frac{3}{4}})\end{cases}\\F=4\int y^\frac{1}{4}dx=4xy^\frac{1}{4}+\phi(y)\\\frac{dF}{dy}=\frac{x}{y^\frac{3}{4}}+\phi'(y)\\\frac{x}{y^\frac{3}{4}}+\phi'(y)=-(104y^\frac{9}{4}-\frac{x}{y^\frac{3}{4}})\\\phi'(y)=-104y^\frac{9}{4}\\\phi(y)=-104\int y^\frac{9}{4}dy=-32y^\frac{13}{4}+C\\4xy^\frac{1}{4}-32y^\frac{13}{4}+C=0\\y(8)=1\\4*8-32+C=0\\C=0\\OTBET:4xy^\frac{1}{4}-32y^\frac{13}{4}=0$