2018高等數(shù)學(xué)輔導(dǎo)講義練習(xí)題詳解-第四章解答

第四 常微分方 【解】應(yīng)選(D).由y 1 1lnyarctanxCy(0)知Cln，故lnyarctanxlnx1 lnyarctan1lnlnye4，故選4則其特征方程為(r1)2r1)r3r2r10，故應(yīng)選（B）.【解】應(yīng)選(A).r210r1,2iyax2bxcx(AsinxBcos【解】應(yīng)選(D).由yCexCe2xxex為方程的,r1,r2為兩個(gè)特征根 特征方程為(r1)(r2)r2r20,正確選項(xiàng)只可能是(C)或(Dyxex代入(D) (r1)(r24)r3r24r40,故應(yīng)選【解】應(yīng)選(C).特征方程為r220,則 ,則特解形式y(tǒng)x(aexbex故應(yīng)選dy

xln y (lnx ln

1 x【解】原方程即y ，解此一階線性方程，得yx

x C

x 6知C1，故所求方程特解為y1x3 xx

(1

ey)

x)y u,則

uydu u 1 y ， du 1 u lnueulnC1u

即(ueuy1

1xyey1Cy1(exexxex.yayby0yCexC yyexyaxexa1則方程yyex的通 yCexCex

1 y(0)0,y(0)3得C1,C1y1(exexxex 1【解】應(yīng)填yC1cosxC2sinxx2xsinxr210,r1,2i.yC1cosxC2sinyaxx[bcosxcsin1代入原方程得a1,b0,c limy(xxlimy(x)1f(0)yx1yx2yexx0 y(0)2.a【解】特征方程r24r30,r11r23,ae2x代入原方程a2,yCe3xCex2e2x. r32r2r20，即(r2)(r21)0r2,iyC1e2xC2cosxC2sinxyC(5)t5(t ytC2t5tt22tYCexCe2x yAxexA2y(xCexCe2x2xex x x x x即C1C21,C12C21.解得C11C20y12x)ex

y3y1exyCx2Cex3 yCx2Cex 式求導(dǎo) y2C1xC2 y2C1 (12

yy2C1(1yyC1(x22x)

聯(lián)立（5）式和（4）式消去C1(2xx2)y(x22)y2(1x)y6(1

1,y

(

1

( xxe2y，則其通解xCeyCey1e2 【解】f(x)x2xf(t)dt xt2f(t)dtx2 f(x)2xxf(t)dtx2f(t)x2f(x)02x[f(x)f(0)]2xf(0)0f(x2xf(x)2x f(x)e2xdx2xe2xdxdxCex22xex2dxex2(ex2C)1Cex22將f(0)0代入上式，得C1，所 f(x)ex xtu則0tf(xt)dt0(xuf(u)x0f(u)du0uf 0f(t)dtxx0f(u)du0uf(u)duxx0f(x)10

f(u)du,f(x)

f(xf(0)1f(x x 【解】對(duì)原方程兩邊求導(dǎo)：

(x)e

[f(t)]0

e[f(t)] f(x)f(x)ex[f(x)]2f f2(x)f(x)e令u1，則u(x)f

uuf

f2(x2由求解公式得u(xedx[edx(ex)dxCex[e2xdxC]Cex1ex21 f(x) 1Cex12 C又由原方程知f(0)1，代入上式得1f(0) 1,C2.所以f(x)3exexC2【解】（1）由題設(shè)知(x1)f(x)(x1)f(x) xf(t)dt0，上式兩邊對(duì)x求導(dǎo)0 (x1)f(x)(x2)f x

Cex設(shè)u

uf(x)u x

xf(0)

ex

f

1，從而

1f

x【證法一】當(dāng)x0時(shí)，f(x1，即f(x)f(0)1f(x)f(0)1設(shè)(x)f(x)ex，則(0)0,(x)f(x)ex x

exx0(x)0，即(x)單調(diào)增加，因而(x)(0)0，即 f(x)ex.x0時(shí)，成立不等式exf(xx x【證法二】由于0f(t)dtf(xf(0)f(x1，所以f(x)10t1x x 注意到當(dāng)x0時(shí)，00t1dt0edt1 ，因而 f(x)1f(t)

(x2y2)f(x2y2)dxdyt42dtr3f(r)drtx2y2t 2tr3f(r)drt4f(t2t3f(t4t3f(00,f(x1(ex4 0【解】f(x)

f(xx)f

exf(x)exf(x)fexlimf(x)f exf(0)f (f(0)2exf(xf(x)x2y2t則ztf(tzz(t) 2z2z(t)4x2z 2z 2 2z 4yz y)z 4z tz(t)z(t) ln 解得z(t)lnt,f(t) ,te最大，f(e) 【解】由題設(shè)知[exf(xff(xCCexxexf(0)4,f(0)3得C0,C4.f(x)4ex 1du[exf(x)]ydxf(x)dyf(x)ydxf(x)dyydf(x)f(x)dyu(x,y)yf(x)Cy(4x)exC.1yy(xP(xy處的法線方程為YyyXx令Y0,XyyxPQ的中點(diǎn)為yy2

x,2

)，由于中點(diǎn)位于拋物線2

xyyx2y)2由此解得

e2x2x 【解】1）解線性方 xf(x)f(x)3ax2得2由21(Cx3ax2dxC4a

f(x)Cx32 f(x)(4a)x

322）V(a)1[(4a)x3ax2]2 π(16a1a2 a5時(shí)，VM的坐標(biāo)為(xyMA的方程為YyyXX0，則YyxyA的坐標(biāo)為(0,yxy由

，有yxy (x0)2(yy2yy1yxz(x0)2(yyx1dx 1

解 zex

xdxCx(xC

y2x2由于所求曲線在第一象限內(nèi)，故y Cxx2 再以條件y33代入得C3.于是曲線方程為y 3xx2(0x yy(x在(xy處的法線方程是Yy

1(Xx)

(y((yy)2y(1y2)2 （y0也滿足上式）. x1y1,y0yPyPdPyPdP1 dPdy 1 y1P0y1P2y2dyPyy2

dxy2積分上式，并注意到時(shí)x1時(shí)y1，得ln(y y21)(xy2y2因此，所求曲線方程為y e(x1)， y1(ex1e(x1))ch(y22【解】(1LP(xy的切線方程為YyyXxX0x2x2線在y軸上的截距為yxy.由題設(shè) yxy，令u

x x2 x2

1 由L經(jīng)過(guò)點(diǎn)1,0，知C1.于是L方程為y 1，即y1x2 x2

（2）y1x2P(xy4Y1x22x(Xx)，即Y2xXx2 0x1.

x2

2 1 ,0與0,x .所求面積

41S(x)

0 x求導(dǎo)，得S(x)

1 1 4 4

1x2

13x2

1 4x2

4 4S(x)0，解得x

3.當(dāng)0x6

3S(x0x36

336x 3S(x在01 2Y23X31

Y

3X1 【解】當(dāng)x0時(shí)，設(shè)(xy k1 .由題意，法線斜率為，所以有 x2y2 2222

，得Cy 2x2, x0. 當(dāng)0xyyx0的通解為yC1cosxC2sinxx yC1sinxC2cosx1. 因?yàn)榍€yy(x)光滑，所以y(x)連續(xù)且可導(dǎo)，由①式知22x2y(0)limy(x) ，22x2x x x 代入②，③式，得C1,C21，故ycosxsinx 0x因 y(x)

2x2 xcosxsinx 0x sin sin【解】曲線L的切線斜率k fycostsint(xffy0xx0f(tcostfsin cost

f(t)

cos2t sintf(t0f(t)