三常微分方程答案

1．SKIPIF1<0=2xy,并满足初始条件：x=0,y=1的特解。解：SKIPIF1<0=2xdx两边积分有：ln|y|=xSKIPIF1<0+cy=eSKIPIF1<0+eSKIPIF1<0=cexSKIPIF1<0SKIPIF1<0另外y=0也是原方程的解，c=0时，y=0原方程的通解为y=cexSKIPIF1<0,x=0y=1时c=1特解为y=eSKIPIF1<0.2.ySKIPIF1<0dx+(x+1)dy=0并求满足初始条件：x=0,y=1的特解。解：ySKIPIF1<0dx=-(x+1)dySKIPIF1<0dy=-SKIPIF1<0dx两边积分:-SKIPIF1<0=-ln|x+1|+ln|c|y=SKIPIF1<0另外y=0,x=-1也是原方程的解x=0,y=1时c=e特解：y=SKIPIF1<03．SKIPIF1<0=SKIPIF1<0解：原方程为：SKIPIF1<0=SKIPIF1<0SKIPIF1<0SKIPIF1<0dy=SKIPIF1<0dx两边积分：x(1+xSKIPIF1<0)(1+ySKIPIF1<0)=cxSKIPIF1<04.(1+x)ydx+(1-y)xdy=0解：原方程为：SKIPIF1<0dy=-SKIPIF1<0dx两边积分：ln|xy|+x-y=c另外x=0,y=0也是原方程的解。5．（y+x）dy+(x-y)dx=0解：原方程为：SKIPIF1<0=-SKIPIF1<0令SKIPIF1<0=u则SKIPIF1<0=u+xSKIPIF1<0代入有：-SKIPIF1<0du=SKIPIF1<0dxln(uSKIPIF1<0+1)xSKIPIF1<0=c-2arctgu即ln(ySKIPIF1<0+xSKIPIF1<0)=c-2arctgSKIPIF1<0.6.xSKIPIF1<0-y+SKIPIF1<0=0解：原方程为：SKIPIF1<0=SKIPIF1<0+SKIPIF1<0-SKIPIF1<0则令SKIPIF1<0=uSKIPIF1<0=u+xSKIPIF1<0SKIPIF1<0du=sgnxSKIPIF1<0dxarcsinSKIPIF1<0=sgnxln|x|+c7.tgydx-ctgxdy=0解:原方程为：SKIPIF1<0=SKIPIF1<0两边积分：ln|siny|=-ln|cosx|-ln|c|siny=SKIPIF1<0=SKIPIF1<0另外y=0也是原方程的解，而c=0时，y=0.所以原方程的通解为sinycosx=c.8SKIPIF1<0+SKIPIF1<0=0解：原方程为：SKIPIF1<0=SKIPIF1<0eSKIPIF1<02eSKIPIF1<0-3eSKIPIF1<0=c.9.x(lnx-lny)dy-ydx=0解：原方程为：SKIPIF1<0=SKIPIF1<0lnSKIPIF1<0令SKIPIF1<0=u,则SKIPIF1<0=u+xSKIPIF1<0u+xSKIPIF1<0=ulnuln(lnu-1)=-ln|cx|1+lnSKIPIF1<0=cy.10.SKIPIF1<0=eSKIPIF1<0解：原方程为：SKIPIF1<0=eSKIPIF1<0eSKIPIF1<0eSKIPIF1<0=ceSKIPIF1<011SKIPIF1<0=(x+y)SKIPIF1<0解：令x+y=u,则SKIPIF1<0=SKIPIF1<0-1SKIPIF1<0-1=uSKIPIF1<0SKIPIF1<0du=dxarctgu=x+carctg(x+y)=x+c12.SKIPIF1<0=SKIPIF1<0解：令x+y=u,则SKIPIF1<0=SKIPIF1<0-1SKIPIF1<0-1=SKIPIF1<0u-arctgu=x+cy-arctg(x+y)=c.13.SKIPIF1<0=SKIPIF1<0解:原方程为：（x-2y+1）dy=(2x-y+1)dxxdy+ydx-(2y-1)dy-(2x+1)dx=0dxy-d(ySKIPIF1<0-y)-dxSKIPIF1<0+x=cxy-ySKIPIF1<0+y-xSKIPIF1<0-x=c14:SKIPIF1<0=SKIPIF1<0解：原方程为：（x-y-2）dy=(x-y+5)dxxdy+ydx-(y+2)dy-(x+5)dx=0dxy-d(SKIPIF1<0ySKIPIF1<0+2y)-d(SKIPIF1<0xSKIPIF1<0+5x)=0ySKIPIF1<0+4y+xSKIPIF1<0+10x-2xy=c.15:SKIPIF1<0=(x+1)SKIPIF1<0+(4y+1)SKIPIF1<0+8xySKIPIF1<0解：原方程为：SKIPIF1<0=（x+4y）SKIPIF1<0+3令x+4y=u则SKIPIF1<0=SKIPIF1<0SKIPIF1<0-SKIPIF1<0SKIPIF1<0SKIPIF1<0-SKIPIF1<0=uSKIPIF1<0+3SKIPIF1<0=4uSKIPIF1<0+13u=SKIPIF1<0tg(6x+c)-1tg(6x+c)=SKIPIF1<0(x+4y+1).16:证明方程SKIPIF1<0SKIPIF1<0=f(xy),经变换xy=u可化为变量分离方程，并由此求下列方程：y(1+xSKIPIF1<0ySKIPIF1<0)dx=xdySKIPIF1<0SKIPIF1<0=SKIPIF1<0证明：令xy=u,则xSKIPIF1<0+y=SKIPIF1<0则SKIPIF1<0=SKIPIF1<0SKIPIF1<0-SKIPIF1<0，有：SKIPIF1<0SKIPIF1<0=f(u)+1SKIPIF1<0du=SKIPIF1<0dx所以原方程可化为变量分离方程。令xy=u则SKIPIF1<0=SKIPIF1<0SKIPIF1<0-SKIPIF1<0(1)原方程可化为：SKIPIF1<0=SKIPIF1<0[1+（xy）SKIPIF1<0](2)将1代入2式有：SKIPIF1<0SKIPIF1<0-SKIPIF1<0=SKIPIF1<0(1+uSKIPIF1<0)u=SKIPIF1<0+cx17.求一曲线，使它的切线坐标轴间的部分初切点分成相等的部分。解：设（x+y）为所求曲线上任意一点，则切线方程为：y=y’(x-x)+y则与x轴，y轴交点分别为：x=xSKIPIF1<0-SKIPIF1<0y=ySKIPIF1<0-xSKIPIF1<0y’则x=2xSKIPIF1<0=xSKIPIF1<0-SKIPIF1<0所以xy=c18.求曲线上任意一点切线与该点的向径夹角为0的曲线方程，其中SKIPIF1<0=SKIPIF1<0。解：由题意得：y’=SKIPIF1<0SKIPIF1<0dy=SKIPIF1<0dxln|y|=ln|xc|y=cx.SKIPIF1<0=SKIPIF1<0则y=tgSKIPIF1<0x所以c=1y=x.19.证明曲线上的切线的斜率与切点的横坐标成正比的曲线是抛物线。证明：设(x,y)为所求曲线上的任意一点，则y’=kx则：y=kxSKIPIF1<0+c即为所求。常微分方程习题2.11.SKIPIF1<0,并求满足初始条件：x=0,y=1的特解.解：对原式进行变量分离得SKIPIF1<0SKIPIF1<0并求满足初始条件：x=0,y=1的特解.解：对原式进行变量分离得：SKIPIF1<03SKIPIF1<0解：原式可化为：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<012．SKIPIF1<0解SKIPIF1<0SKIPIF1<0SKIPIF1<015．SKIPIF1<0SKIPIF1<016．SKIPIF1<0解：SKIPIF1<0SKIPIF1<0，这是齐次方程，令SKIPIF1<017.SKIPIF1<0解：原方程化为SKIPIF1<0令SKIPIF1<0方程组SKIPIF1<0SKIPIF1<0则有SKIPIF1<0令SKIPIF1<0当SKIPIF1<0当SKIPIF1<0另外SKIPIF1<0SKIPIF1<019.已知f(x)SKIPIF1<0.解：设f(x)=y,则原方程化为SKIPIF1<0两边求导得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<020.求具有性质x(t+s)=SKIPIF1<0的函数x(t),已知x’(0)存在。解：令t=s=0x(0)=SKIPIF1<0=SKIPIF1<0若x(0)SKIPIF1<00得xSKIPIF1<0=-1矛盾。所以x(0)=0.x’(t)=SKIPIF1<0)SKIPIF1<0SKIPIF1<0两边积分得arctgx(t)=x’(0)t+c所以x(t)=tg[x’(0)t+c]当t=0时x(0)=0故c=0所以x(t)=tg[x’(0)t]求下列方程的解1．SKIPIF1<0=SKIPIF1<0解：y=eSKIPIF1<0(SKIPIF1<0eSKIPIF1<0SKIPIF1<0)=eSKIPIF1<0[-SKIPIF1<0eSKIPIF1<0(SKIPIF1<0)+c]=ceSKIPIF1<0-SKIPIF1<0(SKIPIF1<0)是原方程的解。2．SKIPIF1<0+3x=eSKIPIF1<0解：原方程可化为：SKIPIF1<0=-3x+eSKIPIF1<0所以：x=eSKIPIF1<0(SKIPIF1<0eSKIPIF1<0eSKIPIF1<0SKIPIF1<0SKIPIF1<0)=eSKIPIF1<0(SKIPIF1<0eSKIPIF1<0+c)=ceSKIPIF1<0+SKIPIF1<0eSKIPIF1<0是原方程的解。3．SKIPIF1<0=-sSKIPIF1<0+SKIPIF1<0SKIPIF1<0解：s=eSKIPIF1<0(SKIPIF1<0eSKIPIF1<0SKIPIF1<0)=eSKIPIF1<0(SKIPIF1<0)=eSKIPIF1<0(SKIPIF1<0)=SKIPIF1<0是原方程的解。4．SKIPIF1<0SKIPIF1<0，n为常数.解：原方程可化为：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0是原方程的解.5．SKIPIF1<0+SKIPIF1<0=SKIPIF1<0解：原方程可化为：SKIPIF1<0=-SKIPIF1<0SKIPIF1<0(SKIPIF1<0)SKIPIF1<0SKIPIF1<0=SKIPIF1<0是原方程的解.6．SKIPIF1<0SKIPIF1<0解：SKIPIF1<0SKIPIF1<0=SKIPIF1<0+SKIPIF1<0令SKIPIF1<0SKIPIF1<0则SKIPIF1<0SKIPIF1<0=uSKIPIF1<0因此：SKIPIF1<0=SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（*）将SKIPIF1<0SKIPIF1<0带入（*）中得：SKIPIF1<0是原方程的解.SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<013SKIPIF1<0这是n=-1时的伯努利方程。两边同除以SKIPIF1<0，SKIPIF1<0令SKIPIF1<0SKIPIF1<0SKIPIF1<0P(x)=SKIPIF1<0Q(x)=-1由一阶线性方程的求解公式SKIPIF1<0=SKIPIF1<0SKIPIF1<014SKIPIF1<0两边同乘以SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0SKIPIF1<0这是n=2时的伯努利方程。两边同除以SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0SKIPIF1<0P（x）=SKIPIF1<0Q(x)=SKIPIF1<0由一阶线性方程的求解公式SKIPIF1<0=SKIPIF1<0=SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<015SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0这是n=3时的伯努利方程。SKIPIF1<0两边同除以SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0SKIPIF1<0=SKIPIF1<0P(y)=-2yQ(y)=SKIPIF1<0由一阶线性方程的求解公式SKIPIF1<0=SKIPIF1<0=SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<016y=SKIPIF1<0+SKIPIF1<0SKIPIF1<0SKIPIF1<0P(x)=1Q(x)=SKIPIF1<0由一阶线性方程的求解公式SKIPIF1<0=SKIPIF1<0=SKIPIF1<0SKIPIF1<0c=1y=SKIPIF1<0设函数SKIPIF1<0(t)于SKIPIF1<0∞<t<SKIPIF1<0∞上连续，SKIPIF1<0(0)存在且满足关系式SKIPIF1<0(t+s)=SKIPIF1<0(t)SKIPIF1<0(s)试求此函数。令t=s=0得SKIPIF1<0(0+0)=SKIPIF1<0(0)SKIPIF1<0(0)即SKIPIF1<0(0)=SKIPIF1<0故SKIPIF1<0或SKIPIF1<0（1）当SKIPIF1<0时SKIPIF1<0即SKIPIF1<0SKIPIF1<0∞,SKIPIF1<0∞)(2)当SKIPIF1<0时SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0于是SKIPIF1<0变量分离得SKIPIF1<0积分SKIPIF1<0由于SKIPIF1<0，即t=0时SKIPIF1<01=SKIPIF1<0SKIPIF1<0c=1故SKIPIF1<020.试证：（1）一阶非齐线性方程（2.28）的任两解之差必为相应的齐线性方程（2.3）之解；（2）若SKIPIF1<0是（2.3）的非零解，而SKIPIF1<0是（2.28）的解，则方程（2.28）的通解可表为SKIPIF1<0，其中SKIPIF1<0为任意常数.（3）方程（2.3）任一解的常数倍或任两解之和（或差）仍是方程（2.3）的解.证明：SKIPIF1<0（2.28）SKIPIF1<0SKIPIF1<0（2.3）设SKIPIF1<0，SKIPIF1<0是（2.28）的任意两个解则SKIPIF1<0SKIPIF1<0（1）SKIPIF1<0（2）（1）-（2）得SKIPIF1<0即SKIPIF1<0是满足方程（2.3）所以，命题成立。由题意得：SKIPIF1<0（3）SKIPIF1<0（4）1）先证SKIPIF1<0是（2.28）的一个解。于是SKIPIF1<0得SKIPIF1<0SKIPIF1<0故SKIPIF1<0是（2.28）的一个解。2）现证方程（4）的任一解都可写成SKIPIF1<0的形式设SKIPIF1<0是(2.28)的一个解则SKIPIF1<0（4’）于是（4’）-（4）得SKIPIF1<0从而SKIPIF1<0即SKIPIF1<0所以，命题成立。设SKIPIF1<0，SKIPIF1<0是（2.3）的任意两个解则SKIPIF1<0（5）SKIPIF1<0（6）于是（5）SKIPIF1<0得SKIPIF1<0即SKIPIF1<0其中SKIPIF1<0为任意常数也就是SKIPIF1<0满足方程（2.3）（5）SKIPIF1<0（6）得SKIPIF1<0即SKIPIF1<0也就是SKIPIF1<0满足方程（2.3）所以命题成立。21.试建立分别具有下列性质的曲线所满足的微分方程并求解。曲线上任一点的切线的纵截距等于切点横坐标的平方；曲线上任一点的切线的纵截距是切点横坐标和纵坐标的等差中项；解：设SKIPIF1<0为曲线上的任一点，则过SKIPIF1<0点曲线的切线方程为SKIPIF1<0从而此切线与两坐标轴的交点坐标为SKIPIF1<0即横截距为SKIPIF1<0，纵截距为SKIPIF1<0。由题意得：（5）SKIPIF1<0方程变形为SKIPIF1<0SKIPIF1<0于是SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所以，方程的通解为SKIPIF1<0。（6）SKIPIF1<0方程变形为SKIPIF1<0SKIPIF1<0于是SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所以，方程的通解为SKIPIF1<0。22．求解下列方程。（1）SKIPIF1<0解：SKIPIF1<0SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0P(x)=SKIPIF1<0Q(x)=SKIPIF1<0由一阶线性方程的求解公式SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<01、验证下列方程是恰当方程，并求出方程的解。1.SKIPIF1<0解：SKIPIF1<0，SKIPIF1<0=1.则SKIPIF1<0所以此方程是恰当方程。凑微分，SKIPIF1<0得：SKIPIF1<02．SKIPIF1<0解：SKIPIF1<0，SKIPIF1<0.则SKIPIF1<0.所以此方程为恰当方程。凑微分，SKIPIF1<0得SKIPIF1<03．SKIPIF1<0解：SKIPIF1<0SKIPIF1<0则SKIPIF1<0.因此此方程是恰当方程。SKIPIF1<0（1）SKIPIF1<0（2）对（1）做SKIPIF1<0的积分，则SKIPIF1<0=SKIPIF1<0SKIPIF1<0（3）对（3）做SKIPIF1<0的积分，则SKIPIF1<0=SKIPIF1<0=SKIPIF1<0则SKIPIF1<0SKIPIF1<0SKIPIF1<0故此方程的通解为SKIPIF1<04、SKIPIF1<0解：SKIPIF1<0，SKIPIF1<0.SKIPIF1<0.则此方程为恰当方程。凑微分，SKIPIF1<0SKIPIF1<0得：SKIPIF1<05.(SKIPIF1<0sinSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0+1)dx+(SKIPIF1<0cosSKIPIF1<0-SKIPIF1<0sinSKIPIF1<0+SKIPIF1<0)dy=0解：M=SKIPIF1<0sinSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0+1N=SKIPIF1<0cosSKIPIF1<0-SKIPIF1<0sinSKIPIF1<0+SKIPIF1<0SKIPIF1<0=-SKIPIF1<0sinSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0+SKIPIF1<0sinSKIPIF1<0SKIPIF1<0=-SKIPIF1<0sinSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0-SKIPIF1<0cosSKIPIF1<0+SKIPIF1<0sinSKIPIF1<0所以，SKIPIF1<0=SKIPIF1<0，故原方程为恰当方程因为SKIPIF1<0sinSKIPIF1<0dx-SKIPIF1<0cosSKIPIF1<0dx+dx+SKIPIF1<0cosSKIPIF1<0dy-SKIPIF1<0sinSKIPIF1<0dy+SKIPIF1<0dy=0d(-cosSKIPIF1<0)+d(sinSKIPIF1<0)+dx+d(-SKIPIF1<0)=0所以，d(sinSKIPIF1<0-cosSKIPIF1<0+x-SKIPIF1<0)=0故所求的解为sinSKIPIF1<0-cosSKIPIF1<0+x-SKIPIF1<0=C求下列方程的解：6．2x(ySKIPIF1<0-1)dx+SKIPIF1<0dy=0解：SKIPIF1<0=2xSKIPIF1<0,SKIPIF1<0=2xSKIPIF1<0所以，SKIPIF1<0=SKIPIF1<0，故原方程为恰当方程又2xySKIPIF1<0dx-2xdx+SKIPIF1<0dy=0所以，d(ySKIPIF1<0-xSKIPIF1<0)=0故所求的解为ySKIPIF1<0-xSKIPIF1<0=C7.(eSKIPIF1<0+3ySKIPIF1<0)dx+2xydy=0解：eSKIPIF1<0dx+3ySKIPIF1<0dx+2xydy=0eSKIPIF1<0xSKIPIF1<0dx+3xSKIPIF1<0ySKIPIF1<0dx+2xSKIPIF1<0ydy=0所以，deSKIPIF1<0(xSKIPIF1<0-2x+2)+d(xSKIPIF1<0ySKIPIF1<0)=0即d[eSKIPIF1<0(xSKIPIF1<0-2x+2)+xSKIPIF1<0ySKIPIF1<0]=0故方程的解为eSKIPIF1<0(xSKIPIF1<0-2x+2)+xSKIPIF1<0ySKIPIF1<0=C8.2xydx+(xSKIPIF1<0+1)dy=0解：2xydx+xSKIPIF1<0dy+dy=0d(xSKIPIF1<0y)+dy=0即d(xSKIPIF1<0y+y)=0故方程的解为xSKIPIF1<0y+y=C9、SKIPIF1<0解：两边同除以SKIPIF1<0得SKIPIF1<0即，SKIPIF1<0故方程的通解为SKIPIF1<010、SKIPIF1<0解：方程可化为：SKIPIF1<0即，SKIPIF1<0故方程的通解为：SKIPIF1<0即：SKIPIF1<0同时，y=0也是方程的解。11、SKIPIF1<0解：方程可化为：SKIPIF1<0SKIPIF1<0即：SKIPIF1<0故方程的通解为：SKIPIF1<012、SKIPIF1<0解：方程可化为：SKIPIF1<0SKIPIF1<0故方程的通解为：SKIPIF1<0即：SKIPIF1<013、SKIPIF1<0解：这里SKIPIF1<0，SKIPIF1<0SKIPIF1<0方程有积分因子SKIPIF1<0两边乘以SKIPIF1<0得：方程SKIPIF1<0是恰当方程故方程的通解为：SKIPIF1<0SKIPIF1<0即：SKIPIF1<014、SKIPIF1<0解：这里SKIPIF1<0因为SKIPIF1<0故方程的通解为：SKIPIF1<0即：SKIPIF1<015、SKIPIF1<0解：这里SKIPIF1<0SKIPIF1<0SKIPIF1<0方程有积分因子：SKIPIF1<0两边乘以SKIPIF1<0得：方程SKIPIF1<0为恰当方程故通解为：SKIPIF1<0即：SKIPIF1<016、SKIPIF1<0解：两边同乘以SKIPIF1<0得：SKIPIF1<0SKIPIF1<0故方程的通解为：SKIPIF1<017、试导出方程SKIPIF1<0具有形为SKIPIF1<0和SKIPIF1<0的积分因子的充要条件。解：若方程具有SKIPIF1<0为积分因子，SKIPIF1<0（SKIPIF1<0是连续可导）SKIPIF1<0SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0方程有积分因子SKIPIF1<0的充要条件是：SKIPIF1<0是SKIPIF1<0的函数，此时，积分因子为SKIPIF1<0.SKIPIF1<0令SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0此时的积分因子为SKIPIF1<018.设SKIPIF1<0及SKIPIF1<0连续,试证方程SKIPIF1<0为线性方程的充要条件是它有仅依赖于SKIPIF1<0的积分因子.证:必要性若该方程为线性方程,则有SKIPIF1<0,此方程有积分因子SKIPIF1<0,SKIPIF1<0只与SKIPIF1<0有关.充分性若该方程有只与SKIPIF1<0有关的积分因子SKIPIF1<0.则SKIPIF1<0为恰当方程,从而SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.其中SKIPIF1<0.于是方程可化为SKIPIF1<0即方程为一阶线性方程.20.设函数f(u)，g(u)连续、可微且f(u)SKIPIF1<0g(u),\，试证方程yf(xy)dx+xg(xy)dy=0有积分因子u=(xy[f(xy)-g(xy)])SKIPIF1<0证：在方程yf(xy)dx+xg(xy)dy=0两边同乘以u得：uyf(xy)dx+uxg(xy)dy=0则SKIPIF1<0=uf+uySKIPIF1<0+yfSKIPIF1<0=SKIPIF1<0+SKIPIF1<0-yfSKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0而SKIPIF1<0=ug+uxSKIPIF1<0+xgSKIPIF1<0=SKIPIF1<0+SKIPIF1<0-xgSKIPIF1<0=SKIPIF1<0=SKIPIF1<0故SKIPIF1<0=SKIPIF1<0，所以u是方程得一个积分因子21．假设方程（2.43）中得函数M（x,y）N(x,y)满足关系SKIPIF1<0=Nf(x)-Mg(y),其中f(x),g(y)分别为x和y得连续函数，试证方程（2.43）有积分因子u=exp(SKIPIF1<0+SKIPIF1<0)证明：M(x,y)dx+N(x,y)dy=0即证SKIPIF1<0SKIPIF1<0uSKIPIF1<0+MSKIPIF1<0=uSKIPIF1<0+NSKIPIF1<0SKIPIF1<0u(SKIPIF1<0-SKIPIF1<0)=NSKIPIF1<0-MSKIPIF1<0SKIPIF1<0u(SKIPIF1<0-SKIPIF1<0)=NeSKIPIF1<0f(x)-MeSKIPIF1<0g(y)SKIPIF1<0u(SKIPIF1<0-SKIPIF1<0)=eSKIPIF1<0(Nf(x)-Mg(y))由已知条件上式恒成立，故原命题得证。22、求出伯努利方程的积分因子.解：已知伯努利方程为：SKIPIF1<0两边同乘以SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0线性方程有积分因子：SKIPIF1<0，故原方程的积分因子为：SKIPIF1<0，证毕！23、设SKIPIF1<0是方程SKIPIF1<0的积分因子，从而求得可微函数SKIPIF1<0，使得SKIPIF1<0试证SKIPIF1<0SKIPIF1<0也是方程SKIPIF1<0的积分因子的充要条件是SKIPIF1<0其中SKIPIF1<0是SKIPIF1<0的可微函数。证明：若SKIPIF1<0，则SKIPIF1<0又SKIPIF1<0即SKIPIF1<0为SKIPIF1<0的一个积分因子。24、设SKIPIF1<0是方程SKIPIF1<0的两个积分因子，且SKIPIF1<0常数，求证SKIPIF1<0（任意常数）是方程SKIPIF1<0的通解。证明：因为SKIPIF1<0是方程SKIPIF1<0的积分因子所以SKIPIF1<0SKIPIF1<0为恰当方程即SKIPIF1<0，SKIPIF1<0下面只需证SKIPIF1<0的全微分沿方程恒为零事实上：SKIPIF1<0即当SKIPIF1<0时，SKIPIF1<0是方程的解。证毕！求解下列方程1、SKIPIF1<0解：令SKIPIF1<0，则SKIPIF1<0，从而SKIPIF1<0，于是求得方程参数形式得通解为SKIPIF1<0.2、SKIPIF1<0解：令SKIPIF1<0，则SKIPIF1<0，即SKIPIF1<0，从而SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，于是求得方程参数形式得通解为SKIPIF1<0.3、SKIPIF1<0解：令SKIPIF1<0，则SKIPIF1<0，从而SKIPIF1<0SKIPIF1<0=SKIPIF1<0SKIPIF1<0，于是求得方程参数形式的通解为SKIPIF1<0,另外，y=0也是方程的解.4、SKIPIF1<0,SKIPIF1<0为常数解：令SKIPIF1<0，则SKIPIF1<0，从而SKIPIF1<0SKIPIF1<0SKIPIF1<0，于是求得方程参数形式的通解为SKIPIF1<0.5、SKIPIF1<01解：令SKIPIF1<0，则SKIPIF1<0，从而SKIPIF1<0SKIPIF1<0SKIPIF1<0，于是求得方程参数形式的通解为SKIPIF1<0.6、SKIPIF1<0解：令SKIPIF1<0，则SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，从而SKIPIF1<0，于是求得方程参数形式的通解为SKIPIF1<0，因此方程的通解为SKIPIF1<0.2．SKIPIF1<0SKIPIF1<0解：两边同除以SKIPIF1<0，得：SKIPIF1<0SKIPIF1<0即SKIPIF1<04．SKIPIF1<0解：两边同除以SKIPIF1<0，得SKIPIF1<0令SKIPIF1<0则SKIPIF1<0即SKIPIF1<0SKIPIF1<0得到SKIPIF1<0，即SKIPIF1<0另外SKIPIF1<0也是方程的解。6．SKIPIF1<0解：SKIPIF1<0SKIPIF1<0得到SKIPIF1<0即SKIPIF1<0另外SKIPIF1<0也是方程的解。8.SKIPIF1<0解：令SKIPIF1<0则：SKIPIF1<0即SKIPIF1<0得到SKIPIF1<0故SKIPIF1<0即SKIPIF1<0另外SKIPIF1<0也是方程的解。10．SKIPIF1<0解：令SKIPIF1<0即SKIPIF1<0而SKIPIF1<0故两边积分得到SKIPIF1<0因此原方程的解为SKIPIF1<0，SKIPIF1<0。12.SKIPIF1<0解：SKIPIF1<0SKIPIF1<0令SKIPIF1<0则SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0故方程的解为SKIPIF1<014．SKIPIF1<0解：令SKIPIF1<0则SKIPIF1<0那么SKIPIF1<0SKIPIF1<0求得：SKIPIF1<0故方程的解为SKIPIF1<0或可写为SKIPIF1<016．SKIPIF1<0解：令SKIPIF1<0则SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0即方程的解为SKIPIF1<018．SKIPIF1<0解：将方程变形后得SKIPIF1<0SKIPIF1<0同除以SKIPIF1<0得：SKIPIF1<0令SKIPIF1<0则SKIPIF1<0SKIPIF1<0即原方程的解为SKIPIF1<019.X(SKIPIF1<0解：方程可化为2y(SKIPIF1<0令SKIPIF1<0SKIPIF1<027.SKIPIF1<0解：令SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，两边积分得SKIPIF1<0即为方程的通解。另外，SKIPIF1<0，即SKIPIF1<0也是方程的解。28.SKIPIF1<0解：两边同除以SKIPIF1<0，方程可化为：SKIPIF1<0令SKIPIF1<0，则SKIPIF1<0即SKIPIF1<0，SKIPIF1<0SKIPIF1<0两边积分得SKIPIF1<0即SKIPIF1<0为方程的解。29.SKIPIF1<0解：令SKIPIF1<0，则SKIPIF1<0，SKIPIF1<0，那么SKIPIF1<0即SKIPIF1<0两边积分得SKIPIF1<0即为方程的解。30.SKIPIF1<0解：SKIPIF1<0方程可化为SKIPIF1<0SKIPIF1<0两边积分得SKIPIF1<0即SKIPIF1<0为方程的解。31.SKIPIF1<0解：方程可化为SKIPIF1<0两边同除以SKIPIF1<0，得SKIPIF1<0即SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0即SKIPIF1<0两边积分得SKIPIF1<0将SKIPIF1<0代入得，SKIPIF1<0即SKIPIF1<0故SKIPIF1<032.SKIPIF1<0解：方程可化为SKIPIF1<0两边同加上SKIPIF1<0，得SKIPIF1<0（*）再由SKIPIF1<0，可知SKIPIF1<0（**）将（*）/（**）得SKIPIF1<0即SKIPIF1<0整理得SKIPIF1<0两边积分得SKIPIF1<0即SKIPIF1<0另外，SKIPIF1<0也是方程的解。求一曲线，使其切线在纵轴上之截距等于切点的横坐标。解：设SKIPIF1<0为所求曲线上的任一点，则在SKIPIF1<0点的切线SKIPIF1<0在SKIPIF1<0轴上的截距为：SKIPIF1<0由题意得SKIPIF1<0即SKIPIF1<0也即SKIPIF1<0两边同除以SKIPIF1<0，得SKIPIF1<0即SKIPIF1<0即SKIPIF1<0为方程的解。摩托艇以5米/秒的速度在静水运动，全速时停止了发动机，过了20秒钟后，艇的速度减至SKIPIF1<0米/秒。确定发动机停止2分钟后艇的速度。假定水的阻力与艇的运动速度成正比例。解：SKIPIF1<0，又SKIPIF1<0，由此SKIPIF1<0即SKIPIF1<0其中SKIPIF1<0，解之得SKIPIF1<0又SKIPIF1<0时，SKIPIF1<0；SKIPIF1<0时，SKIPIF1<0。故得SKIPIF1<0，SKIPIF1<0从而方程可化为SKIPIF1<0当SKIPIF1<0时，有SKIPIF1<0米/秒即为所求的确定发动机停止2分钟后艇的速度。35.一质量为m的质点作直线运动，从速度等于零的时刻起，有一个和时间成正比（比例系数为k1）的力作用在它上面，此质点又受到介质的阻力，这阻力和速度成正比（比例系数为k2）。试求此质点的速度与时间的关系。解：由物理知识得：SKIPIF1<0根据题意：SKIPIF1<0故：SKIPIF1<0即：SKIPIF1<0(*)式为一阶非齐线性方程，根据其求解公式有SKIPIF1<0SKIPIF1<0又当t=0时，V=0，故c=SKIPIF1<0因此，此质点的速度与时间的关系为：SKIPIF1<036.解下列的黎卡提方程（1）