求下列微分方程的通解

習(xí)題124求以下微分方程的通解dyyexdx解yedx(exedxdxC)ex(exexdxC)ex(xC)(2)xyyx23x2解原方程變?yōu)閥1yx32xxye1dx1dxx[(x32)exdxC]x1[(x32)xdxC]1[(x23x2)dxC]xxx1(1x33x22xC)1x23x2Cx3232x(3)yycosxesinx解yecosdx(esinxecosxdxdxC)esinx(esinxesinxdxC)esinx(xC)(4)yytanxsin2x解yetanxdx(sin2xetanxdxdxC)elncosx(sin2xelncosxdxC)cosx(2sinxcosx1dxC)cosxcosx(2cosx+C)Ccosx2cos2x(5)(x21)y2xycosx0解原方程變形為y2xycosxx21x21ye2xdx2xdxx21(cosxex21dxC)x211[cosx(x21)dxC]1(sinxC)x21x21x21(6)d32d解e3d(2e3ddC)e3(2e3dC)e3(2e3C)2Ce333dy2xy4xdx解ye2xdx(4xe2xdxdxC)ex2(4xex2dxC)ex2x2C)2Cex2(2e(8)ylnydx(xlny)dy0解原方程變形為dx1x1dyylnyyxe1dy1dydyC)ylny(1eylnyy1(1lnydyC)lnyy1(1ln2yC)1lnyClny22lny(9)(x2)dyy2(x2)3dx解原方程變形為dy1y2(x2)2dxx2y1dx2(x2)2e1dxex2[x2dxC]21(x2)[2(x2)dxC](x2)[(x2)2C](x2)3C(x2)(y26x)dy2y0dx解原方程變形為dx3x1ydyy23dy3dyxey[(1y)eydyC]2y3(1231y(

1yy3dyC)C)1y2Cy32求以下微分方程滿足所給初始條件的特解dyytanxsecxy|x00dx解yetanxdx(secxetanxdxdxC)1(secxcosxdxC)1(xC)cosxcosx由y|x00得C0故所求特解為yxsecx(2)dyysinx1dxxxx1sinx1解yexdx(exdxdxC)x1(sinxxdxC)1(cosxC)xxx由y|1得C1故所求特解為y1(1cosx)x(3)dyycotx5ecosxy|4dxx2解yecotxdx(5ecosxecotxdxdxC)1(5ecosxsinxdxC)1(5ecosxC)sinxsinx由y|4得C1故所求特解為y1(5ecosx1)x2sinx(4)dy3y8y|x02dx解ye3dx(8e3dxdxC)e3x(8e3xdxC)e3x(8e3xC)8Ce3x33由y|2得C2故所求特解為y2(4e3x)x0dy23x2y1y|x10dx3x解ye23x2dx23x2dxdxC)x3(1ex31111x3ex2(1ex2dxC)x3ex2(1ex2C)x3211311得Cx2由y|0故所求特解為ye)1x2e23求一曲線的方程這曲線經(jīng)過(guò)原點(diǎn)并且它在點(diǎn)(xy)處的切線斜率等于2xy解由題意知y2xy并且y|x00由通解公式得yedx(2xedxdxC)ex(2xexdxC)ex(2xex2exC)Cex2x2由y|x00得C2故所求曲線的方程為y2(exx1)4設(shè)有一質(zhì)量為m的質(zhì)點(diǎn)作直線運(yùn)動(dòng)趕忙度等于零的時(shí)辰起有一個(gè)與運(yùn)動(dòng)方向一至、大小與時(shí)間成正比(比率系數(shù)為k1)的力作用于它其他還受一與速度成正比(比率系數(shù)為k2)的阻力作用求質(zhì)點(diǎn)運(yùn)動(dòng)的速度與時(shí)間的函數(shù)關(guān)系dvdvk2k1解由牛頓定律Fma得mdtk1tk2v即dtmvmt由通解公式得k2dtk2dtk2tk2tvem(k1temdtC)em(k1temdtC)mmk2tkk2tkmk2tem(1tem1emC)k2k22由題意當(dāng)t0時(shí)v0于是得Ck1m因此k22k2tkk2tkmk2tkmvem(1tem1em1)k2k22k22vk1tk1m(1k2t即em)k2k225設(shè)有一個(gè)由電阻R10、電感L2h(亨)和電源電壓E20sin5tV(伏)串通組成的電路開(kāi)關(guān)K合上后電路中有電源經(jīng)過(guò)求電流i與時(shí)間t的函數(shù)關(guān)系解由回路電壓定律知20sin5t2di10i0即di5i10sin5tdtdt由通解公式得ie5dt(10sin5te5dtdtC)sin5tcos5tCe5t因?yàn)楫?dāng)t0時(shí)i0因此C1因此isin5tcos5te5te5t2sin(5t)(A)46設(shè)曲Lyf(x)dx[2xf(x)x2]dy在右半平面(x0)內(nèi)與路徑?jīng)]關(guān)其中f(x)可導(dǎo)且f(1)1求f(x)解因?yàn)楫?dāng)x0時(shí)所給積分與路徑?jīng)]關(guān)因此[yf(x)][2xf(x)x2]yx即f(x)2f(x)2xf(x)2x或f(x)1f(x)12x1dx1dx1(xdxC)2xC因此f(x)e2x(1e2xdxC)x3x由f(1)1可得C1故f(x)2x1x333求以下伯努利方程的通解(1)dyyy2(cosxsinx)dx解原方程可變形為1dy1cosxsinxd(y1)1sinxcosxy2dxy即dxyy1edx(sinxdxdxC][cosx)eex[(cosxsinx)exdxC]Cexsinx原方程的通解為1Cexsinxydy3xyxy2dx解原方程可變形為1dy3x1x即d(y1)3xy1xy2dxydxy1e3xdx[(x)e3xdxdxC]3x23x2e2(xe2dxC)3x23x23x2e2(1e2C)Ce23

13原方程的通解為13x2yCe2

13(3)dy1y1(12x)y4dx33解原方程可變形為1dy111(12x)即d(y3)y32x1y4dx3y33dxy3edx[(2x1)edxdxC]ex[(2x1)exdxC]2x1Cex原方程的通解為1Cex2x1y3dyyxy5dx解原方程可變形為1dy1x即d(y4)4y44xy5dxy4dxy4e4dx[(4x)e4dxdxC]e4(4xe4xdxC)1Ce4x4原方程的通解為1x1Ce4xy44(5)xdy[yxy3(1lnx)]dx0解原方程可變形為1dy11(1lnx)d(y2)2y22(1lnx)y3dxxy2即dxx22y2exdx[2(1lnx)exdxdxC]12[2(1lnx)x2dxC]xC2xlnx4xx239原方程的通解為1C2xlnx4xy2x2398考據(jù)形如yf(xy)dxxg(xy)dy0的微分方程可經(jīng)變量代換vxy化為可分別變量的方程并求其通解解原方程可變形為dyyf(xy)dxxg(xy)在代換vxy下原方程化為xdvvvf(v)dxx2x2g(v)即g(v)1dxv[g(v)f(v)]dux積分得g(v)dulnxCv[g(v)f(v)]對(duì)上式求出積分后將vxy代回即得通解9用合適的變量代換將以下方程化為可分別變量的方程然后求出通解dy(xy)2dx解令uxy則原方程化為du1u2即dxdudx1u2兩邊積分得arctanuC將uxy代入上式得原方程的通解arctan(xy)C即yxtan(xC)dy11(2)dxxy解令uxy則原方程化為1du11即dxududxu兩邊積分得1u2C12將uxy代入上式得原方程的通解x1(xy)2C即(xy)22xC(C2C1)21(3)xyyy(lnxlny)解令uxy則原方程化為x(1duuuulnu即1dx1duxdxx2)xxxulnu兩邊積分得lnxlnClnlnu即ueCx將uxy代入上式得原方程的通解xyeCx即y1eCxx(4)yy22(sinx1)ysin2x2sinxcosx1解原方程變形為(ysinx1)2cosx令uysinx1則原方程化為du2即1dxcosxucosxu2dudx兩邊積分得xCu將uysinx1代入上式得原方程的通解1xC即y1sinx1ysinx1xC(5)y(xy1)dxx(1xyx2y2)d