求下列微分方程满足所给初始条件的特解:
(1)(y+3)dx+cotxdy=0,y∣x=0=1;
(2)y′sin2=ylny,y∣x=π/2=e;
(3)cosydx+(1+e-x)sinydy=0,y∣x=0=π/4;
(4)y′=x/y+y/x,y∣x=1=2;
(5)dy/dx+3y=8,y∣x=0=2.
(1)dy/(y+3)=-(dx/cotx)=-tanxdx 两边积分得: ln(y+3)=lncosx+c ∴y=Ccosx-3 把y∣x=0=1代入得: C=4 ∴y=4cosx-3 (2)dy/ylnx=dx/sin2x=[(-sin2x-cos2x)/sin2x]dx 两边积分得: In(lny)=-Ccotx ∴lny=Ce-cotx 把y∣x=π/2=e代入得:C=1 ∴lny=e-cotx (3)dx/(1+ex)=-(siny/cosy)dy [ex/(ex+1)]dx=dcosy/cosy 两边积分得: ln(1+ex)=ln(cosy) ∴1+ex=Ccosy 把y∣x=0 =π/4代入得: C=2√2 ∴1+ex=2√2cosy (4)令y/x=u,则y′=u′(x)x+u(x) ∴u′(x)x+u(x)=u+1/u ∴ (dy/dx)x=1/u 即:udu=dx/x; 两边积分:(1/2)u2-In∣x∣+C 把u=y/x代入:y2=Cx2(ln ∣x∣+2) 把y∣x=1=2代入得:C=2 ∴y2=2x2(In∣x∣+2) (5)解dy/dx=-3y得: (1/y)dy=-3dx ∴In∣y∣=-3x+C 积分得:y=Ce-3x 运用常数变量法y=C(x)e-x -3C(x)e-3x+C′(x)e-3x+3C(x)e-3x=8 ∴C′(x)=8e3x ∴C(x)=_8/3)e-3x ∴y=8/3+ce-3x 把y∣x=0 =2代入得:y=(2/3)(4-e-3x)