Question

Найти решение дифференциального уравнения, допускающего понижение порядка.
1)$x^{3} *y'' +x^{2} y'=1$
2)$y*y''+ (y')^{2}=0, y(0)=1, y'(0)=1$

Answer 1

$1)\ \ x^3\cdot y''+x^2\cdot y'=1\\\\y'=z(x)\ \ ,\ \ y''=z'(x)\\\\x^3\cdot z'+x^2\cdot z=1\ \ ,\ \ \ z'+\dfrac{z}{x}=\dfrac{1}{x^3}\ \ ,\ \ \ z=uv\ ,\ z'=u'vz=uv'\ \ ,\\\\u'v+uv'+\dfrac{uv}{x}=\dfrac{1}{x^3}\ \ ,\ \ \ u'v+u\, (v'+\dfrac{v}{x})=\dfrac{1}{x^3}\ \ ,\\\\a)\ \ \dfrac{dv}{dx}=-\dfrac{v}{x}\ \ ,\ \ \int \dfrac{dv}{v}=-\int \dfrac{dx}{x}\ \ ,\ \ ln|v|=-ln|x|\ \ ,\ \ v=\dfrac{1}{x}$

$b)\ \ \dfrac{du}{dx}\cdot \dfrac{1}{x}=\dfrac{1}{x^3}\ \ ,\ \ \int du=\int \dfrac{dx}{x^2}\ \ ,\ \ u=-\dfrac{1}{x}+C_1\\\\c)\ \ z=\dfrac{1}{x}\cdot (-\dfrac{1}{x}+C_1)=-\dfrac{1}{x^2}+\dfrac{C_1}{x}\\\\d)\ \ y'=-\dfrac{1}{x^2}+\dfrac{C_1}{x}\ \ ,\ \ \int dy=\int (-\dfrac{1}{x^2}+\dfrac{C_1}{x})\, dx\ \ ,\\\\y=\dfrac{1}{x}+C_1\cdot ln|x|+C_2$

$2)\ \ y\cdot y''+(y')^2=0\ \ ,\ \ y(0)=1\ ,\ y'(0)=1\\\\y'=p(y)\ \ ,\ \ y''=\dfrac{dp}{dy}\cdot p\\\\y\cdot \dfrac{dp}{dy}\cdot p+p^2=0\ \ ,\ \ \dfrac{dp}{dy}\cdot yp=-p^2\ \ \int \dfrac{dp}{p}=-\int \dfrac{dy}{y}\ \ ,\\\\ln|p|=-ln|y|+lnC_1\ \ ,\ \ \ p=\dfrac{C_1}{y}\ \ ,\ \ y'=\dfrac{C_1}{y}\ \ ,\\\\y'(0)=1\ ,\ y(0)=1:\ \ 1=\dfrac{C_1}{1}\ \ \to \ \ C_1=1\\\\y'=\dfrac{1}{y}\ \ ,\ \ \int y\cdot dy=\int dx\ ,\\\\\dfrac{y^2}{2}=x+C_2\ \ ,\ \ y^2=2\, (x+C_2)\ \ ,$

$y(0)=1:\ \ 1=2(0+C_2)\ \ \to \ \ C_2=\dfrac{1}{2}\\\\y^2=2\, (x+\dfrac{1}{2})\ \ ,\ \ \ \underline {\ y^2=2x+1\ }$