
解答:u=y′⇒u′−3u=8e3x+4⇒integration factor I(x)=e∫−3dx=e−3x⇒u′e−3x−3e−3xu=8+4e−3x⇒(e−3xu)′=8+4e−3x⇒e−3xu=8x−43e−3x+c1⇒u=y′=8xe3x−43+c1e3x⇒y=83xe3x−89e3x−43x+13c1e3x+c2⇒y=83xe3x−43x+c3e3x+c2
解答:{u(x,y,z)=2x2+3y2+z2⇒∇u=(ux,uy,uz)=(4x,6y,2z)v=→i−2→k=(1,0,−2)⇒→v‖→v‖=1√5(1,0,−2)⇒D→vu(2,1,3)=∇u(2,1,3)⋅→v‖→v‖=(8,6,6)⋅1√5(1,0,−2)=−4√5

解答:(a)→r(θ,ϕ)=(2cosϕsinθ,2sinϕsinθ,2cosθ)⇒→n=→rθ×→rϕ=(4sin2θcosϕ,4sin2θsinϕ,4sinθcosθ)⇒∬S(→F⋅→n)dA=∫πθ=0∫2πϕ=0(14sinθcosϕ,0,−2cosθ)⋅(4sin2θcosϕ,4sin2θsinϕ,4sinθcosθ)dθ=∫πθ=0∫2πϕ=056cos2ϕsin2θ−8cos2θsinθdϕdθ=64π(b)→F=7x→i−z→k⇒div→F=Fx+Fy+Fz=7+0−1=6⇒∬S(→F⋅→n)dA=∭Rdiv→FdV=∭R6dV=6⋅43π⋅23=64π
解答:
令u(x,t)=F(x)G(t),則{utt=FG″uxx=F″G,代回原式可得FG″=c2F″G⇒G″c2G=F″F由於G″c2G只含變數t,而F″F只含變數x,兩者等值代表同為一常數k,即G″c2G=F″F=k則邊界條件:{u(0,t)=F(0)G(t)=0u(L,t)=F(L)G(t)=0,若G(t)=0,則u(x,t)=0為明顯解,不列入討論因此邊界條件轉變成F(0)=F(L)=0;F″−kF=0⇒F=c1e√kx+c2e−√kx,代入初始值⇒{F(0)=c1+c2=0⋯(1)F(L)=c1e√kL+c2e−√kL=0⋯(2)由(1)得c2=−c1代入(2)⇒c1(e√kL−e−√kL)=0⇒c1=0或e√kL=e−√kL若c1=0⇒c2=0⇒F=0為明顯解,不列入討論;因此只需考慮e√kL=e−√kL而e√kL≠e−√kL,∀k>0;若k=0⇒F=0為明顯解,不列入討論因此只有k<0,此時F(x)=c1(ei√|k|x−e−i√|k|x)=2ic1sin(√|k|x)⇒F(L)=2ic1sin(√|k|L)=0⇒√|k|L=nπ,n∈N⇒√|k|=nπ/L⇒F(x)=2ic1sin(nπx/L)=Asin(nπLx),A為常數k=−n2π2L2⇒G″+n2π2L2c2G=0⇒G(t)=c3cos(nπLct)+c4sin(nπLct)因此u(x,t)=Asin(nπLx)(c3cos(nπLct)+c4sin(nπLct))=∞∑n=1sin(nπLx)(Ancos(nπLct)+Bnsin(nπLct))
初始條件1:u(x,0)=f(x)⇒f(x)=∞∑n=1Ansin(nπLx)⇒∫L0f(x)sin(kπLx)dx=∞∑n=1∫L0Ansin(kπLx)sin(nπLx)dx=∫L0Aksin2(kπLx)dx=L2Ak⇒Ak=2L∫L0f(x)sin(kπLx)dx⇒An=2L∫L0f(x)sin(nπLx)dx=2L(∫L/202kLxsin(nπLx)dx+∫LL/22kL(L−x)sin(nπLx)dx)⇒An=8knπ(1nπsin(nπ2)−cos(nπ2)+cos(nπ))
初始條件2:∂u∂t|t=0=0⇒∞∑n=1(−cnπLAnsin(cnπLt)+cnπLBncos(cnπLt))sin(nπLx)|t=0=∞∑n=1cnπLBnsin(nπLx)=0⇒Bn=0⇒u(x,t)=∞∑n=1sin(nπLx)(Ancos(nπLct))⇒u(x,t)=∞∑n=18knπ(1nπsin(nπ2)−cos(nπ2)+cos(nπ))sin(nπLx)cos(nπLct)
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解題僅供參考,其他歷年試題及詳解
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