求通解，最好是手写答案，我要详细点，谢谢 求定积分，最好手写谢谢
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解法1： 令 y/x = p， 则 y = xp, dy/dx = p+xdp/dx
p+xdp/dx = 1-p, dp/(1-2p) = dx/x
-(1/2) ln(1-2p) = lnx - (1/2)LnC, 1/(1-2p) = x^2/C
x^2(1-2y/x) = C, x(x-2y) = C
解法2： dy/dx + y/x = 1
y = e^(∫-dx/x) [∫e^(∫dx/x)dx - C/2]
= (1/x) [ ∫xdx - C/2] = (1/x) [ (1/2)x^2 - C/2]
2xy = x^2 - C, x^2-2xy = C
绛旓細dy/dx = 1 - y/x 瑙ｆ硶1锛 浠 y/x = p锛 鍒 y = xp, dy/dx = p+xdp/dx p+xdp/dx = 1-p, dp/(1-2p) = dx/x -(1/2) ln(1-2p) = lnx - (1/2)LnC, 1/(1-2p) = x^2/C x^2(1-2y/x) = C, x(x-2y) = C 瑙ｆ硶2锛 dy/dx + y/x =...
绛旓細鍥炵瓟锛氳y鈥=u;鍒欏師鏂圭▼u鈥=u^3+u;u=u^4/4+u^2/2+C1
绛旓細寰垎鏂圭▼y''-y'=e^x鐨勯氳В涓簓=Ce^x+De^(-x)+0.5xe^x銆傝В绛旇繃绋嬪涓嬶細y''-y=0鐨勭壒寰佹柟绋嬩负a^2-1=0 瑙ｆ槸a=1鎴朼=-1 鍥犳閫氳В鏄y=Ce^x+De^(-x)銆倅''-y=e^x鐨勭壒瑙ｈ涓簓=e^x(ax)鍒檡'=ae^x(x+1)锛寉''=ae^x(x+2)浠ｅ叆鏂圭▼寰2ae^x=e^x 浜庢槸a=0.5锛岀壒瑙ｆ槸y...
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绛旓細鍥炵瓟锛歽"=-x-y .... y'=-1/2 x²-xy+c1 .... y=-1/6 x³-1/2 x²y+c1x+c2 濂藉儚鏄繖鏍峰惂
绛旓細=x arctanx -½ ln(1+x²) +C1 x +C2 (2)浠'=p锛寉''=p'鍒檟p'+p=0 dp/p=-dx/x ln|p|=-ln|x|+ln|C| 鍗硃=C/x y'=C/x y=C ln|x|+C1 锛3锛変护y'=p锛寉''=p'鍒檖'=1+p²dp/(1+p²)=dx arctan p=x+c p=tan(x+c)y'=tan(x+...
绛旓細鍏堟眰榻愭閫氾紝鍐嶆眰闈為綈鐗 绛旀濡傚浘鎵绀
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