台湾大学研究生入学-微分方程习题解答与参考

Solve

x2y’-2xy-y=0,y(1)=e-1(2009台大生醫電資所)

(Sol.),,,y(1)=e-1A=1,∴.

(y2+1)dx=ysec2(x)dy(2002成大電研)

(Sol.),,.

y’=ytan(2x),y(0)=2(2003中央光電所)

(Sol.),ln|y|+c=-ln|cos(2x)|,=cos(2x),y(0)=2A=4,∴=cos(2x).

y”=1+(y’)2(台科大電研)

(Sol.)u=y’,u’=1+u2,,tan-1(u)=x+c,u=y’=tan(x+c),∴y=ln[sec(x+c)]=d.

tf’-(t+2)f=-2t2-2t(2013台大電研)

(Sol.)Itisthefirst-orderlineardifferentialequation.f’-()f=-2t-2,,t-2e-tf’-t-3(t+2)e-tf=(-2)．(t-1+t-2)e-t,[t-2e-tf]’=-2(t-1+t-2)e-t,t-2e-tf=2t-1e-t+c,∴f=2t+ct2et.

y’=y-y2(2009台大生醫電資所)

(Sol.)ItisBernouli’sequation,letz=y1-2=y-1,y=z-1,y2=z-2,

,,,,

[zex]’=ex,zex=ex+c,z=1+ce-x,∴.

(2y2+3x)dx+2ydy=0(台大電研)

(Sol.)y’+y=1.5xy-1,itisBernouli’sequation,letz=y1-(-1)=y2,y=z1/2,y-1=z-1/2,

,z’+2z=,[ze2x]’=,

ze2x=+c,z=+ce-2x,∴y2=+ce-2x.

y’+y/x=xy2

(Sol.)ItisBernouli’sequation,letz=y1-2=y-1,y=z-1,y2=z-2,

,,,

,,,z=-x2+cx,∴y-1=-x2+cx.

.

(Sol.)y’+y/x=2x-3y-4/3,itisBernouli’sequation,letz=y1-(-4/3)=y7/3,y=z3/7,

,,

,,,,

,.

xy’+2y=xy3(1991清大電研)

(Sol.)ItisBernouli’sequation,let

,,

,

x(lny-lnx)dy=(ylny-ylnx-x)dx(2005台大電研)

(Sol.)Itisahomogeneousequation.,

.Letu=y/x,ln(u)[u+xu’]=uln(u)-1,

ln(u)du=-uln(u)-u=-ln(x)+c,∴.

(1991交大電信所)

(Sol.)Set

(2y2-9xy)dx+(3xy-6x2)dy=0(2005台科大電研)

(Sol.)y’=,letu=y/x,,,

,,,

∴.

(1990交大電子所)

(Sol.).Let,,

,ue-udu=dx,-ue-u-e-u=x+c,

(x2+2xy-y2)dx+(y2+2xy-x2)dy=0(2009台大電研)

(2009台灣聯合大學系統碩士班聯招)

(Sol.).Letu=y/x,,,

,∴.

(文化電機轉學考)(Sol.).

(2006台科大電研)

(Sol.)ItisaType-2quasi-homogeneousequation.Letv=2x+y,y=v-2x,y’=v’-2=,v’=+2=,=dx,()dv=dx,=x+c,

∴=x+c.

y’+=0(2005成大電研)

(Sol.)y’=.ItisaType-1quasi-homogeneousequation.LetAandBfulfill-2A+5B+9=0,4A-B-9=0A=2,B=-1x=X+2,y=Y-1,Y’==.Letu=Y/X,,=,

du=,+=,-2ln(u+2)+ln(u-1)=ln(X)+c,

ln[]=ln(AX),AX=,A(x-2)=,

∴A(x-2)=.

(2006台科大電研)

(Sol.),

=,∴xy+xy2=C

(2x+ey)dx+(xey)dy=0(2003中央光電所)

(Sol.)Itistheexactdifferentialequation,∴f(

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