高数 08.06 微分方程习题课 02

微分方程习题课 02
(一)单选题
1.下面各微分方程中为一阶线性方程的是(\ \ {\color{blue}{B} }\ \ )
A.xy′+y2=xB.y′+xy=sinxC.yy′=xD.(y′)2+xy=0
解:Adydx+1xy2=1,不是Bdydx+xy=sinx,是

2.微分方程(x+y)y′+(x−y)=0的通解是(  B  )
A.12(x2+y2)=CearcsinyxB.arctanyx+lnx2+y2−−−−−−√=CC.ln(x2+y2)=arctanyx+CD.x2+y2−−−−−−√=Cearctanyx
解:(x+y)y′+(x−y)=0y′=y−xy+x=yx−1yx+1令u=yx,u+u′x=y′=u−1u+1u′x=u−1−u2−uu+1=−1−u2u+1(u+1)duu2+1=−dxx∫(u+1)duu2+1=∫−dxx12∫d(u2+1)u2+1+∫duu2+1=∫−dxx12∫d(u2+1)u2+1+∫dxx=−∫duu2+112ln(u2+1)+lnx=−arcsinu+C12ln(u2+1)x2=−arcsinu+C12ln[(yx)2+1]x2=−arcsinyx+Clnx2+y2−−−−−−√=−arcsinyx+Clnx2+y2−−−−−−√+arcsinyx=+C

3.微分方程y′+yx=1x(x2+1)的通解是(  B  )
A.arctanx+CB.1x(arctanx+C)C.1xarctanx+CD.arctanx+Cx
解:P(x)=1x,Q(x)=1x(x2+1),y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫1xdx[C+∫1x(x2+1)e∫1xdxdx]=e−lnx[C+∫1x(x2+1)elnxdx]=1x[C+∫1x(x2+1)xdx]=1x[C+∫1(x2+1)dx]=1x(C+arctanx)

4.微分方程y′−12(y+1)=0的解是(  D  )
A.y=x2+1B.y=(x+1)2C.x=y2+1D.x=(y+1)2
解:变形得:dxdy=2y+2dx=(2y+2)dy∫dx=∫(2y+2)dyx=y2+2y+C

(二)填空题
1.微分方程d4sdt4+s=s4的自变量为  t  −−−,未知函数为  s  −−−,方程的阶数为  4  −−−

2.微分方程x2y′+y2=0的自变量为  x  −−−,未知函数为  y  −−−,方程的阶数为  1  −−−

3.微分方程(x2+y2)dx−xydy=0为一阶  齐次  −−−−−方程
解:(x2+y2)dx=xydydydx=yx+1yx

4.微分方程dydx−1xy=0的通解是  Cx  −−−−−
解:P(x)=−1x,Q(x)=0;y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫−1xdx[C+∫0e∫−1xdxdx]=elnx[C+0]=Cx

5.微分方程y′+y=0的同解是  y=Ce−x  −−−−−−−−−−
解:dydx=−ydyy=−dx∫dyy=∫−dx−x=lny+lnC1C1y=e−xy=Ce−x

6.微分方程dy−2xdx=0的通解为  y=x2+C  −−−−−−−−−−−,满足y|x=0=0的特解为  y=x2  −−−−−−−
解:dy=2xdxy=x2+C0=02+C,C=0;y=x2

6.微分方程ylnxdx+xlnydy=0的通解为  (lny)2=−(lnx)2+C  −−−−−−−−−−−−−−−−−−−,满足y|x=e12=e12的特解为  y=(lny)2=−(lnx)2+12  −−−−−−−−−−−−−−−−−−−−−−−
解:ylnxdx+xlnydy=0lnydyy=−lnxdxx∫lnydyy=∫−lnxdxx∫lnydlny=∫−lnxdlnx∫d(lny)2=−∫d(lnx)2(lny)2=−(lnx)2+C把y|x=e12=e12代入,C=12特解:(lny)2=−(lnx)2+12

(三)解答题
1.求微分方程(1+y2)dx=−(1+x2)dy的通解
解:dy1+y2=−dx1+x2∫dy1+y2=∫−dx1+x2通解:arctany=−arctanx+C

2.求微分方程dydx=−xy的通解,并求满足初始条件y(2)=4的特解
解:方程变形:ydy=−xdx∫ydy=∫−xdx12∫dy2=−12∫dx2∫dy2=−∫dx2通解:x2+y2+C(C为任意常数)把y(2)=4代入,C=20,特解:x2+y2=20

3.求微分方程xdy+dx=eydx的通解
解:方程变形得:xdy=(ey−1)dxdxx=dyey−1∫dxx=∫dyey−1lnx=∫(1−ey+ey)dyey−1=∫(eyey−1−1)dy=−y+∫eyey−1dy=−y+∫d(ey−1)ey−1=−y+ln(ey−1)+C即:lnx=−y+ln(ey−1)+C

4.求微分方程sinxcosydx=cosxsinydy的通解,并求满足初始条件y(0)=π4的特解
解:方程变形得:tanydy=tanxdx∫tanydy=∫tanxdx∫dln(cosy)=∫dln(cosx)ln(cosy)=ln(cosx)+lnC通解:cosy=Ccosx(C为任意常数)把y(0)=π4代入，得2√2=C⋅1,C=2√2特解:cosy=2√2cosx

5.求微分方程(ex+y−ex)dx+(ex+y−ey)dy=0的通解
解：方程变形得:ex(ey−1)dx=−ey(ex−1)dyeydyey−1=−exdxex−1d(ey−1)ey−1=−d(ex−1)ex−1∫d(ey−1)ey−1=∫−d(ex−1)ex−1ln(ey−1)=−ln(ex−1)+lnCln(ex−1)(ey−1)=lnC通解:(ex−1)(ey−1)=C

例6.求微分方程dydx=yy−x的通解
解:方程变形为:dxdy=1−xy令u=xy,u+u′y=x′=1−uu′y=1−2udu1−2u=dyy∫du1−2u=∫dyy∫dyy=−12∫d(1−2u)1−2u2∫dyy=−∫d(1−2u)1−2u2lny=−ln(1−2u)+lnCy2(1−2u)=Cy2(1−2xy)=Cy2−2xy=C

7.求微分方程xdydx=y(lny−lnx)的通解
解:方程变形得:dydx=yxlnyx令u=yx,u+u′x=y′=ulnuu′x=ulnu−uduu(lnu−1)=dxxd(lnu−1)(lnu−1)=dlnxdln(lnu−1)=dlnx∫dln(lnu−1)=∫dlnxln(lnu−1)=lnx+lnClnu−1=Cxlnu=Cx+1u=eCx+1yx=eCx+1y=xeCx+1(C为任意常数)

8.求微分方程xdydx−y−y2−x2−−−−−−√=0的通解,并求满足初始条件y(1)=1的特解
解:方程变形为:dydx=yx+(yx)2−1−−−−−−−√令u=yx,u+u′x=y′=u+u2−1−−−−−√u′x=u2−1−−−−−√duu2−1−−−−−√=dxx∫duu2−1−−−−−√=∫dxx
参考高数04.02换元法 例3
ln|u+u2−1−−−−−√|=ln|x|+ln|C|u+u2−1−−−−−√=Cxyx+(yx)2−1−−−−−−−√=Cx通解:y+y2−x2−−−−−−√=Cx2(C为任意常数)把y(1)=1代入,C=1,特解:y+y2−x2−−−−−−√=x2

例9.求微分方程(x+y)dydx+(x−y)=0的通解
解:方程变形为:dydx=y−xy+x=yx−1yx+1令u=yx,u+u′x=y′=u−1u+1u′x=u−1−u2−uu+1=−1+u2u+1(u+1)du1+u2=−dxx∫(u+1)du1+u2=−∫dxx12∫d(1+u2)1+u2+∫du1+u2=−∫dlnxln(1+u2)+2arctanu=−lnx2−lnClnCx2(1+u2)+2arctanu=0lnC(x2+y2)+2arctanyx=0

10.求微分方程(x−ycosyx)dx+xcosyxdy=0的通解
解:方程变形得:dydx=−(x−ycosyx)xcosyx=−(1−yxcosyx)cosyx令u=yx,u+u′x=y′=−(1−ucosu)cosuu′x=−1−ucosu+ucosucosu=−1cosucosudu=−dxxdsinu=−dlnx∫dsinu=∫−dlnxsinu=−lnx+C通解:sinyx=−lnx+C

11.求微分方程xy′+y=x2+3x+2的通解
解:方程变形为:y′+1xy