Queen's College, Hong Kong



Differentiation exercise – show differential equation1.If y=x sin 2x , prove that xd2ydx2-2dydx+2yx+4xy=0y=x sin 2x?dydx=sin 2x+2xcos2xd2ydx2=2cos 2x+2cos2x-4xsin2xxd2ydx2-2dydx+2yx+4xy=2x cos 2x+2xcos2x-4x2sin2x-2sin 2x+4xcos2x+2x sin 2xx+4x2sin2x=02.Given that y=ex-e-x , show that dydx2-y2-4=0y=ex-e-x?dydx=ex+e-xdydx2-y2-4=ex+e-x2-ex-e-x2-4=e2x+2+e-2x-e2x-2+e-2x-4=03.Given that v=sinu , show that 4v3d2vdu2+v4+1=0dvdu=cos u2 sin ud2vdu2=12sin u-sinu-cos ucos u2sin usin u=14-2sin2 u-cos2 usin usin u=-sin2 u+14sin usin u=-v4+14v3∴4v3d2vdu2+v4+1=04.Given y=e-xcosx , show that d2ydx2+2dydx+2y=0.Method 1 y=e-xcosx dydx=-e-xsinx -e-xcosx d2ydx2=e-xsinx-e-xcosx +e-xsinx+e-xcosx =2e-xsinx ∴d2ydx2+2dydx+2y=2e-xsinx -2e-xsinx+e-xcosx +2e-xcosx =0Method 2y=e-xcosx ?lny=-x+lncosx1ydydx=-1-sinxcosxdydx=-y-y tan xdydx+2y=y1-tanxDifferentiate, we get: d2ydx2+2dydx=dydx1-tanx-y sec2 x=-y-y tan x1-tanx-y 1+tan2 x=-2y∴d2ydx2+2dydx+2y=0Method 3 y=e-xcosx ?yex=cosx …(1)dydxex+yex=-sinxd2ydx2ex+dydxex+dydxex+yex=-cosx d2ydx2ex+2dydxex+yex=-yex , by (1)∴d2ydx2+2dydx+2y=05.Given that y=sinkx1+coskx , where k is a positive integer, show that sinkx d2ydx2=k2y2.Method 1dydx=1+coskxkcoskx-sinkx-ksinkx1+coskx2=kcoskx+kcos2kx+ksin2kx1+coskx2=kcoskx+k1+coskx2=k1+coskx1+coskx=k1+coskxsinkx dydx=ksinkx1+coskx=kysinkx d2ydx2+kcoskxdydx=kdydx sinkx d2ydx2=kdydx1-coskx=kdydx1-coskx1+coskx1+coskx=kdydxsin2kx1+coskx=ksinkx dydxsinkx1+coskx=kkyy=k2y2Method 2Given that y=sinkx1+coskx , where k is a positive integer, show that sinkx d2ydx2=k2y2.Note that : y=sinkx1+coskx=1-coskxsinkxdydx=sinkxksinkx-1-coskxkcoskxsin2kx=ksin2kx+kcos2kx-kcoskxsin2kx=k-kcoskxsin2kx=ksinkx1-coskxsinkx=kysinkxsinkxdydx=kysinkxd2ydx2+kcos kxdydx=kdydxsinkxd2ydx2=k1-coskxdydx=k1-coskxsinkxsinkxdydx=kyky=k2y26.Given y(2 – x) = 3, show that 3d2ydx2-2ydydx=0.Method 1Differentiate, 2 – xdydx-y=0 dydx=y2-xDifferentiate again,d2ydx2=2-xdydx-y-12-x2=2-xy2-x+y2-x2=2y2-x2 3d2ydx2-2ydydx=32y2-x2-2yy2-x=6y2-x2-2y22-x=6y2-x2-2yy2-x2-x2=6y2-x2-2y32-x2=0Method 2Differentiate, 2 – xdydx-y=0Differentiate again,2 – xd2ydx2-dydx-dydx=0or2 – xd2ydx2-2dydx=0Multiply by y, y(2 – x)d2ydx2-2dydx=0∴3d2ydx2-2ydydx=07.Given y=1+4xe-2x , prove that d2ydx2+4dydx+4y=0Method 1 dydx=2e-2x-8xe-2xd2ydx2=16xe-2x-12e-2xd2ydx2+4dydx+4y=16xe-2x-12e-2x+42e-2x-8xe-2x+41+4x=16xe-2x-12e-2x+8e-2x-32xe-2x+4e-2x+16xe-2x=0Method 2e2xy=1+4xe2xdydx+2ye2x=4e2xd2ydx2+2e2xdydx+2e2xdydx+4e2xy=0e2xd2ydx2+4dydx+4y=0d2ydx2+4dydx+4y=08.Let y=cosx , show that 4y3d2ydx2+y4+1=0.y=cosx?y2=cosxDifferentiate, 2ydydx=-sinx …(1)Differentiate again, 2yd2ydx2+2dydxdydx=-cosx=-y22yd2ydx2+2dydx2=-y2Multiply by 2y2,4y3d2ydx2+2ydydx2=-2y4By (1),4y3d2ydx2+-sinx2=-2y4, 4y3d2ydx2+1-cos2 x=-2y44y3d2ydx2+1-y4=-2y44y3d2ydx2+y4+1=09.Given 1+x2y2=1-x2 , show that dydx2=1-y41-x4 .Method 11+x2y2=1-x2 …(1)y2=1-x21+x2, 1-y2=1-1-x21+x2=2x21+x2, 1+y2=1+1-x21+x2=21+x21-y4=1-y21+y2=4x21+x22Differentiate (1), 2xy2+1+x22ydydx=-2xdydx=-2x1+y22y1+x2, dydx2=4x21+x221+y224y2=1-y421+x2241-x21+x2=1-y41+x21-x2=1-y41-x4Method 2Let x=tanθ2, cosθ=1-x21+x2, dxdθ=121+x2y2=1-x21+x2=cosθ, y=cosθ dydθ=-sinθ2cosθ, dydθ2=sin2θ4cosθ=1-cos2θ4cosθ=1-y441-x21+x2dydx2=dydθ2dxdθ2=1-y441-x21+x2121+x22=1-y41+x21-x2=1-y41-x410.Form a differential equation from y=Ax3+Bx2-6x, x>0 .yx2=Ax5+B-6x3Differentiate, x2dydx+2xy=5Ax4-18x2 Divide by x, xdydx+2y=5Ax3-18x …(1)Differentiate (1), xd2ydx2+dydx+2dydx=15Ax2-18xd2ydx2+3dydx=15Ax2-18Multipy by x, x2d2ydx2+3xdydx=15Ax3-18xx2d2ydx2+3xdydx=35Ax3-18x+36x=3xdydx+2y+36x , by (1)x2d2ydx2-6xy=36xxd2ydx2-6y=3611.Form a differential equation from y=Ax3+Bx2-6, x>0 .yx2=Ax5+B-6x2Differentiate, x2dydx+2xy=5Ax4-12x Divide by x, xdydx+2y=5Ax3-12 …(1)Differentiate (1), xd2ydx2+dydx+2dydx=15Ax2xd2ydx2+3dydx=15Ax2Multipy by x, x2d2ydx2+3xdydx=15Ax3x2d2ydx2+3xdydx=35Ax3-12+36=3xdydx+2y+36 , by (1)x2d2ydx2-6y=36x3d2ydx2-6xy=36x12.y sin-13x=1-9x2 , show that 1-9x2dydx+3y2+9xy=0y sin-13x=1-9x2sin-13xdydx+y31-9x2=-18x21-9x2ysin-13xdydx+y231-9x2=-9xy1-9x21-9x2dydx+3y21-9x2+9xy1-9x2=01-9x2dydx+3y2+9xy=013. Given that y=xnAcoslnx+Bsinlnx , where A and B are constants, show thatx2d2ydx2+1-2nxdydx+1+n2y=0y=xnAcoslnx+Bsinlnx …(1)dydx=nxn-1Acoslnx+Bsinlnx+xn-Axsinlnx+BxcoslnxMultiply by x, we have,xdydx=nxnAcoslnx+Bsinlnx+xn-Asinlnx+BcoslnxBy (1),xdydx=ny+xn-Asinlnx+Bcoslnx …(2)Differentiate again,xd2ydx2+dydx=ndydx+nxn-1-Asinlnx+Bcoslnx-xnAxcoslnx+BxsinlnxMultipy by x, x2d2ydx2+xdydx=nxdydx+nxn-Asinlnx+Bcoslnx-xnAcoslnx+Bsinlnxx2d2ydx2+xdydx=nxdydx+nxdydx-ny-y , by (1) and (2).x2d2ydx2+1-2nxdydx+1+n2y=014.Given that y=sin-1 x , show that 1-x2d2ydx2-xdydx=0y=sin-1 x?siny=x?cosydydx=1 …1Differentiate (1), cosyd2ydx2-sinydydxdydx=0Multiply by cosy , cos2yd2ydx2-sinydydxcosydydx=01-x2d2ydx2-xdydx1=0 , by (1)1-x2d2ydx2-xdydx=015.Let y=5e3-2x+3e-3+2x , show that d2ydx2+4dydx+y=0.Method 1Let y=u+v, where u=5e3-2x, v=3e-3+2x.dudx=53-2e3-2x, d2udx2=53-22e3-2x=57-43e3-2xHence, d2udx2+4dudx+u=57-43e3-2x+203-2e3-2x+5e3-2x=5e3-2x7-43+43-2+1=0dvdx=-53+2e-3+2x, d2vdx2=53+22e-3+2x=57+43e-3+2xHence, d2vdx2+4dvdx+v=57+43e-3+2x-203-2e-3+2x+5e-3+2x=5e-3+2x7+43-43+2+1=0d2ydx2+4dydx+y=d2u+vdx2+4du+vdx+u+v=d2udx2+4dudx+u+d2vdx2+4dvdx+v=0Method 2y=5e3-2x+3e-3+2xe2xy=5e3x+3e-3x …(1)e2xdydx+2e2xy=53e3x-33e-3xe2xd2ydx2+2e2xdydx+2e2xdydx+4e2xy=15e3x+93e-3xe2xd2ydx2+4e2xdydx+4e2xy=3e2xy , by (1)e2xd2ydx2+4e2xdydx+e2xy=0Method 3Let α=3-2, β=-3+2.α+ β=-4, α β=-22-3=1α,β are roots of u2+4u+1=0 …(1)Hence α2+4α+1=0β2+4β+1=0 …(2)y=5e3-2x+3e-3+2x=5eαx+3eβxdydx=5αeαx+3βeβx, d2ydx2=5α2eαx+3β2eβxd2ydx2+4dydx+y=5eαxα2+4α+1+3eβxβ2+4β+1=0 , by (2).Comment(a)The general solution of the differential equation: d2ydx2+4dydx+y=0 isy=c1e3-2x+c2e-3+2x, c1,c2 are integrating constants.So y=5e3-2x+3e-3+2x is a special solution of the differential equation.(b)It is interesting that solving d2ydx2+4dydx+y=0 is just to find the roots of the auxiliary equation u2+4u+1=0, that is α=3-2, β=-3+2 and then form the general solution y=c1e3-2x+c2e-3+2x. Solving this kind of second order differential equation becomes solving a quadratic equation. The story is longer since quadratic equation may have equal roots or complex roots and is not discussed here.(c)Showing differential equation is just learning how to dive in the sea of calculus and solving differential equation is to get some pearls in the sea.19/1/2018Yue Kwok Choy ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download