講高數(shù)17堂專題課微分方程有關(guān)綜合題

PAGEPAGE2專題 微分方程有關(guān)綜合1】f(xf(xxf(x2xf(x)x(xx0f(02, 【解】f(xxf(x2xf(x)x(xx0)f(xx)f(x)2xf(x)上式中令x0

f(x)2xf解方程 f(x)f(0)2,則C2,f(x2ex2,f(1)2】f(x在(,f(0)2xyf(xy)exfy)eyf(xf(x【解 f(x)

f(xx)f

exf(x)exf(x)fexlimf(x)f exf(0)f (f(0)2exff(x)2xe3】yy(xypyqye3xy(0)y(0)

x

ln(1x2

的極限（（A）不存 （B）等于 （C）等于 （D）等于【解】ypyqye3xy(xy(0)1limln(1x2)limx2 x0lim2x x0y(02,故應(yīng)選

x0y y 4】f(x滿足xf(t)dtxtf(xt)dt f(Ⅱ)若f(x)在區(qū)間[0,1上的平均值為1,a的值【解】xtudt 0tf(xt)dtx(xu)f 由題設(shè)條件

x0f(u)du0uf xf(u)duxxf(u)duxuf(u)du 上式兩端求導(dǎo)xf(x)xxf(x)x

f(u)duxf(x)xf(x)2axf(u)du2axf(x)f(x)f(x)ex[2aexdxCexf(0)0,可得C2af(x)2a(1ex(Ⅱ)由題意可 1f(x)dx10 12a(1ex)dx0PAGEPAGE3a2【例5】設(shè)函數(shù)f(x)在(,)上連續(xù)，且滿f(x)xetf(t)dt1(x f【解】f(x)xetf(t)dt1x1)ex兩端同乘ex exf(x)xetf(t)dt1(x 令xetf(t)dt1F(x0F(x)F(x)x1[F2(x)]x2[F2(x)]2(xF2(x)(x1)2F(01F(xx1,xetf(t)dt1x0exf(x)f(x)【例6】設(shè)f(x)為連續(xù)函數(shù)yayf x

y(x，其a是正常數(shù)

f(x)k（k為常數(shù)x0y(x)k(1eaxa【證1（1）原方程的通解y(x)eadxf(x)eaxdxCeax[F(x)F(xf(x)eax的任一原函數(shù).y(0)0，得CF(0)y(x)eax[F(x)F(0)]eaxxf0PAGEPAGE9ax

ax

k （2）y(x)

f0

dt

eatdt

k(1eaxa

x【證2（1）在原方程的兩端同乘以eax，yeaxayeaxf(x)eax從而(yeax)f(x)eax，所以yeax xf(t)eatdt，0y(x) xf(t)eatdt0（2）同證法一【例7】設(shè)函數(shù)f(x)在[0,上可導(dǎo)，f(0)1，且滿足等f(x)f(x) xf(t)dt0x1（1）求導(dǎo)數(shù)f(x【解】（1）由題設(shè)知(x1)f(x)(x1)f(x) xf(t)dt00上式兩邊對x求導(dǎo)(x1)f(x)(x2)f令uf(x)，則 dux2u 解之 f(x)ux

x.f(01f(0)f(00f(01 ex從而C1

(x)x（21】x0f(x0f(xf(0)1設(shè)(x)

f(x)f(0)1f(x)ex，則(0)0,(x) f(x)ex x

exx0時，(x)0，即(x)單調(diào)增加，因(x)(0)0，即 f(x)ex.x0時，成立不等式exf(x1x【證2】由 0

(t)dt

f(xf(0)f(x1x x

f(x)1

x0t

注意到當(dāng)x0時，00t1dt0edt1e 因而exf(x)1.8】y(xy2yky0其中0k（Ⅰ）證明：反常積分 y(x)dx收斂y(0)1y(0)1求y(x)dx0 1k,r211k0k1,r0r0,從而er1xdx與er1xdx收斂 rryCer1xCer2x，其中C和C為任意常數(shù) 由此可知，反常積分 y(x)dx收斂(Ⅱ)由(Ⅰ)知r10,r20,所limy(x)lim[Cer1xCer2x] limy(x)lim[Crer1xrCer2x] 1 y(x)dx[1(y(x)2 k

(y(x)20kz(x2y2)f(x2y

2z2z

f

)滿x

y

【解】x2y2t則ztf(tzz(t)

2z2z(t)4x2z2z2z(t)4y2z y)z 4z tz(t)z(t) 解得z(tlnt,f(tlnttf(t1lntf(t0得tef(t在tef(t在tet2取極大值，又因為tef(t在[0,f(ef(t在[0,1的最大值，f(e) e10】設(shè)函數(shù)u(xyduexf(x)]ydxf(x)dyf具有二階連續(xù)的導(dǎo)數(shù)，且f(0)4,f(0)3,求f(x)及u(x,y).【解】duexf(x)]ydxf(x)dy[exf(x)]f f(x)f(x)ex解方程得f(x)C1C2exxex,由f(0)4,f(0)3,得C10,C24.則f(x)4exxex.du[exf(x)]ydxfyf(x)dxf(x)dyydf(x)f(x)dyd(yf u(x,y)yf(x)y(4【例11】設(shè)f(x)連續(xù),且f(t) (x2y2)fx2y2t

x2y2dxdyt4f【解 d 2t3f()dt0f(xf(xf(0)0,上式兩端對t f(t)e2t3dte2t3dt(4t3)dtC t4 1t t e2 e1

CCe f(0)0知C f(x) (e【例12（I）驗證函

y(x)1 (xyyyex n（II）利用（I）的結(jié)果求冪級數(shù)(3n)!的和函n

y(x)1x3x6x9

x3n，，2y(x)2

x5x8

x3n ，， (3n，，

x7

x3n y(x)x

(3n所yyyex（2）yyyexyyy0其特征方程特征根為

12

2103i.23x3Ye2C1

33 設(shè)非齊次微分方程的特解

yyyyyexA13方程通解

x

y13 yYy*e2C

3xC

3x x0

y(0)1C1 y(0)01C

3C13 2 n由此，得C13,C20.于是冪級數(shù)(3nnY(x)

23e23

x13

(x,13】yy(x是區(qū)間(,,

光滑曲線.當(dāng)x02曲線上任一點處的法線都過原點；當(dāng)0xy(xyyx0.求函數(shù)y(x)的表達(dá)式.【解】x0時，設(shè)(xy為曲線上任一點，由導(dǎo)數(shù)幾何意義，法線斜率為k1dy.y dyx 分離變量，解y

x2y2222222x2當(dāng)0xyyx0

x