微分方程求解---齐次方程

初次接触齐次方程时，搞不清楚"齐次方程"和"齐次线性方程"的区别。两者"齐次"含义不同，请看这里：https://www.zhihu.com/question/374818288

和这里：
https://www.zhihu.com/question/268048924

下面我们用两个例子继续巩固一下齐次微分方程的求解。

例子1

d y d x = y − x 2 + y 2 x \frac {dy}{dx} = \frac{y - \sqrt{x^2+y^2}}{x} dxdy​=xy−x2+y2 ​​

解：
d y d x = y − x 2 + y 2 x = = > d y d x = y x − 1 + ( y x ) 2 \frac {dy}{dx} = \frac{y - \sqrt{x^2+y^2}}{x} ==>\\ \frac {dy}{dx} = \frac{y}{x} - \sqrt{1+ ( \frac{y}{x})^2}\\ dxdy​=xy−x2+y2 ​​==>dxdy​=xy​−1+(xy​)2 ​

设 $y=ux$ , 则 d y = u d x + x d u , d y d x = u + x d u d x dy =udx + xdu, \frac{dy}{dx} = u + x \frac{du}{dx} dy=udx+xdu,dxdy​=u+xdxdu​, 带入上式得：
u + x d u d x = u − 1 + u 2 = = > x d u d x = − 1 + u 2 = = > d u − 1 + u 2 = d x x u + x \frac{du}{dx} = u - \sqrt{1+ u^2} ==>\\ x \frac{du}{dx} = - \sqrt{1+ u^2} ==>\\ \frac{du}{ - \sqrt{1+ u^2} } = \frac{dx}{x}\\ u+xdxdu​=u−1+u2 ​==>xdxdu​=−1+u2 ​==>−1+u2 ​du​=xdx​

设u = tant, 代入上式得：
s e c 2 t d t − s e c t = d x x s e c t d t = d x x \frac{sec^2t dt}{ - sect } = \frac{dx}{x}\\ sect dt = \frac{dx}{x} −sectsec2tdt​=xdx​sectdt=xdx​

例子2

x 2 y ′ + x y = y 2 x^2 y ' +xy = y^2 x2y′+xy=y2

解：
x 2 y ′ + x y = y 2 = = > y ′ + y x − y 2 x 2 = 0 x^2 y ' +xy = y^2 ==>\\ y' + \frac{y}{x} - \frac{y^2}{x^2} = 0 x2y′+xy=y2==>y′+xy​−x2y2​=0

设 $y=ux$ , 则 d y = u d x + x d u , d y d x = u + x d u d x dy =udx + xdu, \frac{dy}{dx} = u + x \frac{du}{dx} dy=udx+xdu,dxdy​=u+xdxdu​, 带入上式得：

u + x d u d x + u − u 2 = 0 = = > d u u 2 − 2 u = d x x = = > u + x\frac{du}{dx} + u - u^2 = 0 ==>\\ \frac {du}{u^2 -2u} = \frac{dx}{x} ==> u+xdxdu​+u−u2=0==>u2−2udu​=xdx​==>

1 2 [ d u u − 2 − d u u ] = d x x = = > 1 2 [ l n ∣ u − 2 ∣ − l n ∣ u ∣ ] = l n ∣ x ∣ \frac{1}{2} [\frac {du}{u-2} - \frac {du}{u} ]= \frac{dx}{x} ==>\\ \frac{1}{2} [ ln|u-2| - ln|u|] = ln|x| 21​[u−2du​−udu​]=xdx​==>21​[ln∣u−2∣−ln∣u∣]=ln∣x∣

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