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1. Solve

i) x2y’-2xy-y=0, y(1)=e-1 (2009台大生醫電資所)

(Sol.) [pic], [pic], [pic], y(1)=e-1[pic]A=1, ∴ [pic].

ii) (y2+1)dx=ysec2(x)dy (2002成大電研)

(Sol.) [pic], [pic], [pic].

iii) y’=ytan(2x), y(0)=2 (2003中央光電所)

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

iv) y”=1+(y’)2 (台科大電研)

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

v) tf’-(t+2)f= -2t2-2t (2013台大電研)

(Sol.) It is the first-order linear differential equation. f’-([pic])f=-2t-2, [pic], 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.

vi) y’=y-y2 (2009台大生醫電資所)

(Sol.) It is Bernouli’s equation, let z=y1-2= y-1, y=z-1, y2=z-2,

[pic],[pic], [pic], [pic],

[zex]’=ex, zex=ex+c, z=1+ce-x, ∴ [pic].

vii) (2y2+3x)dx+2ydy=0 (台大電研)

(Sol.) y’+y=1.5xy -1, it is Bernouli’s equation, let z=y1-(-1)= y2, y=z1/2, y-1= z-1/2,

[pic], z’+2z=[pic], [ze2x]’=[pic],

ze2x=[pic]+c, z=[pic]+ce-2x, ∴ y2=[pic]+ce-2x.

viii) y’+y/x=xy2

(Sol.) It is Bernouli’s equation, let z=y1-2= y-1, y=z -1, y2=z -2,

[pic], [pic], [pic],

[pic], [pic], [pic], z=-x2+cx, ∴ y-1=-x2+cx.

ix) [pic].

(Sol.) y’+y/x=2x-3y-4/3, it is Bernouli’s equation, let z=y1-(-4/3)=y7/3, y=z3/7,

[pic], [pic],

[pic], [pic], [pic], [pic],

[pic], [pic].

x) xy’+2y=xy3 (1991清大電研)

(Sol.) It is Bernouli’s equation, let [pic]

[pic]

[pic],[pic],

[pic], [pic][pic][pic]

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

(Sol.) It is a homogeneous equation. [pic],

[pic]. Let u=y/x, ln(u)[u+xu’]=uln(u)-1,

ln(u)du=-[pic][pic]uln(u)-u=-ln(x)+c, ∴ [pic].

xii) [pic] (1991交大電信所)

(Sol.) Set [pic]

[pic]

xiii) (2y2-9xy)dx+(3xy-6x2)dy=0 (2005台科大電研)

(Sol.) y’=[pic], let u=y/x, [pic], [pic],

[pic], [pic], [pic],

∴ [pic].

xiv) [pic] (1990交大電子所)

(Sol.) [pic]. Let [pic], [pic],

[pic], ue-udu=dx, -ue-u-e-u=x+c, [pic]

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

xvi) [pic] (2009台灣聯合大學系統碩士班聯招)

(Sol.) [pic]. Let u=y/x,[pic], [pic],

[pic], ∴ [pic].

xvii) [pic] (文化電機轉學考) (Sol.) [pic].

xviii) [pic](2006台科大電研)

(Sol.) It is a Type-2 quasi-homogeneous equation. Let v=2x+y, y=v-2x, y’=v’-2=[pic], v’ =[pic]+2=[pic], [pic]=dx, ([pic])dv=dx, [pic]=x+c,

∴ [pic]=x+c.

xix) y’+[pic]=0 (2005成大電研)

(Sol.) y’ =[pic]. It is a Type-1 quasi-homogeneous equation. Let A and B fulfill -2A+5B+9=0, 4A-B-9=0[pic]A=2, B=-1[pic]x=X+2, y=Y-1, Y’ =[pic]=[pic]. Let u=Y/X, [pic], [pic]=[pic],

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

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

∴ A(x-2)=[pic].

xx) [pic](2006台科大電研)

xxi) [pic]

(Sol.) [pic], [pic]

[pic][pic]=[pic], ∴ xy+ xy2=C

xxii) (2x+ey)dx+(xey)dy=0 (2003中央光電所)

(Sol.) It is the exact differential equation, ∴ f(x,y)=x2+xey+c=0.

xxiii) (exsiny+3y)dx+(3x+excosy)dy=0 (交大控制所)

(Sol.) It is the exact differential equation, ∴ f(x,y)=exsiny+3xy+c=0.

xxiv) (2x-5y+3)dx-(2x+4y-6)dy=0 (2005成大電研)

xxv) xy’-4x2y+2yln(y)=0 (交大電研)

xxvi) y”+2y’+y=[pic]. (1990台大電研)

(Sol.) r2+2r+1=0, r=-1, -1, yh=c1e-t+c2te-t. Set yp=[pic],

yp’=[pic], yp”=[pic]

[pic][pic]

[pic][pic]

[pic]

xxvii) y”-2y’+y=x2ex (文化電機轉學考、2005台大電研)

xxviii) y”-2y’+y=ex (2009台大電研) (Sol.) y=c1ex+c2xex+[pic].

xxix) y”+y=(x-1)cos(x) (交大控制所)

xxx) y”+3y’+2y=4x2 (2000台大電研)

xxxi) f"-3f’+2f=4t+e3t, f(0)=-f’(0)=1 (2013台大電研) (Sol.) f(t)=[pic]+2t+3.

xxxii) y”-4y’+4y=(x+1)e2x (2006台大電研)

xxxiii) y”+4y’+4y=(x+3)e-2x, y(0)=2, y’(0)=5. (交大電機控制所甲組)

xxxiv) y”+8y’+16y=8sin(2x)+3e4x (2002台科大電研)

xxxv) y”+4y=x2cos(2x) (交大電子研究所)

xxxvi) y”-3y’+2y=x+e2x (交大電研)

xxxvii) y”+3y’+2y=x2(ex+e-x) (交大電研)

xxxviii) x2y”-3xy’+13y=4+3x

(Sol.) It is Euler’s equation. y=[pic].

xxxix) xy”+4y’=[pic](2004台大電研)

(Sol.) x2y”+4xy’=3ln(x). It is Euler’s equation. y(x)=c1+[pic]+[pic].

xl) (x+1)2y”-(x+1)y’+y=0, y(0)=1, y’(0)=0 (交大控制所)

(Sol.) It is similar to Euler’s equation. Let X=(x+1)[pic]X 2y”-Xy’+y=0. And then let z=ln(X)[pic][pic], y=cez+dzez=cX+dXln(X)=c(x+1)+d(x+1)ln(x+1). y(0)=1, y’(0)=0[pic]c=1, d=-1, ∴ y=x+1-(x+1)ln(x+1).

xli) x2y”-xy’+y=√x (交大電子所)

xlii) x3y’”+5x2y”+7xy’+8y=0 (台大電研)

xliii) x3y’”-2x2y”+5xy’-5y=0 (2004台科大電研)

xliv) x2y”+3xy’+y= x2+2x+3 (2005台科大電子所)

xlv) x2y”-xy’+y=ln(x) (2005台科大電研)

2. Find the integrating factor of (2y2-9xy)dx+(3xy-6x2)dy=0 and solve it.

(Sol.) [pic]

Suppose [pic]

[pic][pic][pic][pic]

[pic]

[pic], ∴ [pic]

3. Apply the parameter variation method to obtain the solutions of

i) y”-y=[pic] (1991台大土木所) (Sol.) yp=[pic]-[pic].

ii) y”+3y’+2y=[pic] (交大電信所)

iii) y”+5y’= xe-xsin(3x) (2001成大電研)

iv) y”-y’-12y=2sinh2(x) (2002成大電研)

4. [pic], where x(0)=0 and y(0)=0. Express y(t) in terms of x(t). (2005北科大電通所)

(Sol.) [pic]. Let u=y-x[pic]u’+u=-3x, u=-3x+3+ce-x, ∴ y= -2x+3+ce-x.

5. Find the differential equation associated with the solution y=C1e3x+C2e2x+C3ex. (成大電研)

6. Obtain the orthogonal trajectories of the curves:

i) x2+2y2=K, K>0 (Sol.) y=Ax2.

ii) r=1-sinθ (Sol.) r=k(1+sinθ).

iii) x2+cy2=4 (交大電子研究所).

7. Solve

i) (x2+1)y”+xy’-y=0 (2006台科大電研)

ii) x2y”+x(0.5+2x)y’+(x-0.5)y=0 (Sol.) [pic].

iii) x2y”+xy+ x2-1=0 (2010台大電研)

iv) (x-1)y”+y’=0 (2007台大電研)

v) y”-(1+x)y=0 (2002台大電研)

vi) x2y”+(6x+x2)y’+xy=x2+2x (台灣聯合大學系統碩士班聯招)

vii) (x-1)y”-xy’+y’=0 (台大機研)

viii) xy”+(x-6)y’-3y=0 (交大資訊研究所)

ix) y”-xy=1 (2005台大電研)

8. (a) Define a function g(t) by g(t)=[pic]. What is the Laplace transform of g’(t)? (b) Let h(t)=[pic], then [pic]=? (2005交大電機群組研).

9. Determine

i) L[2e-3t] and L[t2sinh(t)] (1991交大材料所)

(Sol.) L[2e-3t]=[pic], L[t2sinh(t)]=[pic]=[pic]=[pic]-[pic].

ii) L[e-2tcos(4t)] (Sol.) [pic].

iii) L[t3sin(3t)] (Sol.) [pic].

iv) [pic] (2005台科大電研)

(Sol.) [pic]=[pic]

=e-πs/24．[pic]=e-πs/24．[pic]=[pic].

v) If L[f(t)]=F(s) and k>0, then [pic]=? (2003成大電研)

(Sol.) [pic].

vi) [pic] (1991台大環工所) (Sol.) [pic]=[pic]=e4t+3e-4t.

vii) [pic] (2001台大電研)

(Sol.) [pic]=[pic], ∴[pic]=[pic].

viii) [pic] (1991交大材料所)

(Sol.) [pic]=sin(t), [pic]=[pic]=1-cos(t),

[pic]=[pic]=t-sin(t).

ix) [pic] (2012台大電研)

(Sol.) [pic]=2-[pic]=2-[pic]=2-[pic]-[pic],

[pic]=2δ(t)-cos(√2t)e-t-sin(√2t)e-t.

x) [pic] (2003成大電研)

(Sol.) [pic]=[pic]+[pic],[pic]= cos(2t)e5t-sin(2t)e5t.

xi) [pic] (2006台科大電研)

xii) [pic] (2003成大電研)

(Sol.) [pic]cos(πt)+sin(πt), ∵ [pic],

∴ [pic]={cos[π(t-1)]+sin[π(t-1)]}．u(t-1).

xiii) [pic] (1991清大電研)

(Sol.) [pic]

[pic].

xiv) [pic] (2013台大電研)

(Sol.) [pic]=[pic]．([pic]-[pic])

=[pic]．[[pic]-[pic]], and

[pic]=[pic],[pic]=[pic]

[pic]=[pic]．[pic]=[pic].

xv) [pic] (台大電研)

(Sol.) [pic], ∵ [pic],

∴[pic]=(t-2)e2(t-2)u(t-2).

xvi) [pic] (成大電研)

(Sol.) [pic]=[pic]=[pic], ∴[pic]=δ(t)-3sin(3t).

xvii) [pic] (成大電研)

(Sol.) [pic]=ln(s)+ln(s+1)-2ln(s-2)=[pic]- [pic]+2[pic],

∴ [pic]=[pic].

xviii) [pic] (1991交大材研) (Sol.)[pic]

[pic] [pic][pic][pic]=e0t(0+t)+0+(-1)sin(t)e0t=t-sin(t).

xix) [pic]

(Sol.) [pic],

[pic][pic][pic].

xx) [pic] (台科大電研)

xxi) Find [pic].

(Sol.) According to[pic] and [pic]

[pic][pic][pic]

10. Find [pic].

(Sol.) [pic]

According to [pic] and [pic],

[pic][pic][pic]

Set s=1, [pic].

11. Find [pic].

(Sol.) According to [pic] and [pic],

[pic] , [pic], [pic],

[pic]

According to [pic], ∴ [pic].

12. Find [pic]. (2013中山光電所)

(Sol.) [pic]=[pic]=[pic].

Set s=0, [pic].

13. Eg. Solve y(4)=δ(x-a), y(0)=y”(0)=y(1)=0, y(3)(0)=1, 0 ................
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