Question

求一道高数题20.

Answer 1

由于yy"-y'^2=y^2y',那么(yy"-y'^2)/y^2 =y',注意到y'/y的导数就是(yy"-y'^2)/y^2,所以对等式两边积分得y'/y=y+A (A为常数),那么dy/ [(y+A)*y] =dx,即[1/y -1/(y+A)] *dy=Adx,积分得到ln|y/(y+A)|= Ax+B,(A,B为常数)即|y...

Answer 2

Answer 3

该微分方程属于缺 x 型，即缺自变量型。
设 y' = p 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 y^2pdp/dy = p, p = 0 或 y^2dp/dy = 1
p = y' = 0, 得 y = C；
y^2dp/dy = 1 即 dp = dy/y^2， p = -1/y+C1 = (C1y-1)/y = dy/dx
dx = ydy/(C1y-1) = (1/C1)[1+1/(C1y-1)]dy
x = (1/C1)[y+(1/C1)ln|C1y-1|] + C2

Answer 4

由原来的定积分∫<-1,x>√(1-e^t)dt=y
很明显有积分上限x=-1时，y=0

Answer 5

追问

倒数第二步不明白

追答

分离变量