國立臺北科技大學111學年度碩士班招生考試
系所組別 :1201製 造科技研究所
第一節 微分方程 試題 (選考)
解答:令{P(x,y)=4yQ(x,y)=x+12xy⇒{Py=4Qx=1+12y⇒Py≠Qx⇒Not exact取u′=−Py−QxPu=(3−34y)u⇒u=y−3/4e3y⇒{uP=4y1/4e3yuQ=xy−3/4e3y+12xy1/4e3y⇒(uP)y=(uQ)x=y−3/4e3y+12y1/4e3yΦ(x,y)=∫4y1/4e3ydx=∫xy−3/4e3y+12xy1/4e3ydy⇒4xy1/4e3y+ϕ(y)=4xy1/4e3y+ρ(x)⇒4xy1/4e3y+c1=0解答:1+2xtany+(x2−xtany)y′=0⇒(1+2xtany)dx+(x2−xtany)dy=0令{P(x,y)=1+2xtanyQ(x,y)=x2−xtany⇒Py≠Qx⇒ Not Exact⇒u′=Qx−PyPu=−tanyu⇒u=−c1cosy⇒{uP=−c1cosy(1+2xtany)uQ=−c1cosy(x2−xtany)⇒Φ(x,y)=∫cos(y)(1+2xtany)dx=∫cos(y)(x2−xtany)dy⇒∫cosy+2xsinydx=∫x2cosy−xsinydy⇒xcosy+x2siny+ϕ(y)=x2siny+xcosy+ρ(x)⇒xcosy+x2siny+c1=0y(−1)=π⇒1+c1=0⇒c1=−1⇒xcosy+x2siny=1
解答:y′+y=y2⇒dyy2−y=dx⇒ln(y−1)−lny=x+c1⇒y−1y=c2ex⇒y=11−c2ex⇒y(0)=11−c2=−13⇒c2=4⇒y=11−4ex
解答:先求齊次解,即y‴+3y″+3y′+y=0⇒λ3+3λ2+3λ+1=0⇒(λ+1)3=0⇒λ=−1⇒yh=e−x(c1+c2x+c3x2)特解yp=Ax3e−x⇒y′p=3Ax2e−x−Ax3e−x⇒y″p=6Axe−x−6Ax2e−x+Ax3e−x⇒y‴p=6Ae−x−18Axe−x+9Ax2e−x−Ax3e−x⇒y‴p+3y″p+3y′p+yp=6Ae−x=30e−x⇒A=5⇒yp=5x3e−x⇒y=yh+yp=e−x(c1+c2x+c3x2+5x3)⇒y(0)=c1=3⇒y′=e−x(c2−3+(2c3−c2)x+(15−c3)x2−5x3)⇒y′(0)=c2−3=−3⇒c2=0⇒y″=e−x(2c3+3+(30−4c3)x+(c3−30)x2+5x3)⇒y″(0)=2c3+3=−47⇒c3=−25⇒y=e−x(3−25c2+5x3)
解答:L{y⁗}+3L{y″}−4L{y}=(s4Y(s)−s3y(0)−s2y′(0)−sy″(0)−y‴(0))+3(s2Y(s)−sy(0)−y′(0))−4Y(s)=0⇒(s4+3s2−4)Y(s)=10−10s2⇒Y(s)=−10(s2−1)(s2+4)(s2−1)⇒Y(s)=−10s2+4⇒y(t)=L−1{Y(s)}=−5L−1{2s2+22}=−5sin(2t)⇒y=−5sin(2t)
=========================== END ==============================
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言