Question

y'-(2/x+1)y=(x+1)^2微方方程

Answer 1

求方程dy/dx-2y/x+1=（x+1）^5/2的通解。(1)dy/dx-2y/(x+1)=(x+1)^(5/2)x+1=udx=dudy/du-2y/u=u^(5/2)y/u=sdy=uds+sduuds/du+s-2s=u^(5/2)uds/du-s=u^(5/2)uds-sdu=u^(5/2)du(uds-sdu)/u^2=u^(1/2)dud(s/u)=(2/3)du^(3/2)+Cs/u=(2/3)u^(3/2)+Cy/u^2=(2/3)u^(3/2)+Cy=(2/3)u^(7/2)+Cu^2通解y=(2/3)(x+1)^(7/2)+C(x+1)^2(2)dy/dx–2y/（x+1）=（x+1）^5/2为一阶线性方程,由通解公式：y=(x+1)^2(C+∫(x+1)^(1/2)dx)=(x+1)^2(C+(2/3)(x+1)^(3/2))=C(x+1)^2+(2/3)(x+1)^(7/2)

追问

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