Question

求解(3)(4)(5)(6)题

Answer 1

直接分离变量就可以了！
(3)y'=x√(1-y²)
==> dy/dx=x√(1-y²)
==> dy/√(1-y²)=xdx
==> arcsiny=(1/2)x²+C
==> y=sin[(1/2)x²+C]

(4)ylnxdx+xlnydy=0
==> ylnxdx=-xlnydy
==> [(lnx)/x]dx=-[(lny)/y]dy
==> (1/2)(lnx)²=-(1/2)(lny)²+C1
==> (lnx)²+(lny)²=C

(5)cosxsinydx+sinxcosydy=0
==> cosxsinydx=-sinxcosydy
==> (cosx/sinx)dx=-(cosy/siny)dy
==> (1/sinx)d(sinx)=-(1/siny)d(siny)
==> ln(sinx)=-ln(siny)+C1
==> ln(sinx)+ln(siny)=C1
==> ln(sinxsiny)=C1
==> sinxsiny=C

(6)(y+1)²(dy/dx)+x³=0
==> (y+1)²dy=-x³dx
==> (1/3)(y+1)³=(-1/4)x^4+C1
==> (y+1)³=(-3/4)x^4+C