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Question 1: Find and sketch the largest region of the xy-plane for which the initial value problemx2-x-2dydx= 1-ln?(2-y)y0=0,has a unique solution.Question 2:Solve the initial value problemxy2+4xdx+8y-2x2ydy=0, x<2y0=0Obtain the general solution of the differential equation2cosxdydx+ysinx-4x+52y3=0, x∈(-π2,π2)Question 3:Solve the differential equationcosy+ycosxdx+(sinx-xsiny)dy=0By using an appropriate substitution, or any other method, find the general solution of the differential equationdydx= yxlnxy, x>0, y>0Question 4: Let y(1+cosx)=C, where x∈(0,π), be a given family of curves. Find the family of orthogonal trajectories for this family of curves.Question 1: Find and sketch the largest region of the xy-plane for which the initial value problemx2-x-2dydx= 1-ln?(2-y)y0=0,has a unique solution.Solution:dydx= 1-ln2-yx2-x-2=fx,yf is continuous for: x2-x-2 ≠0, 2-y≥0, 1-ln2-y>0 x2-x ≠2, 2≥y, 1>ln?(2-y) xx-1≠2, y≤2, e>2-y x≠2, x≠-1 2-e<yfx,y=1x2-x-2 . 1-ln2-y1/2 , ?f?y=(1x2-x-2 )121-ln2-y-12 . -1(2-y) ?f?y is continuous for: x2-x-2 ≠0, 2-y≥0, 1-ln2-y≥0, 2-y≠0 x2-x ≠2, 2≥y, 1≥ln?(2-y), y≠2 xx-1≠2, y≤2, e≥2-y151415914224000 x≠2, x≠-1 2-e≤yf and ?f?y, are continuous on: x,y ∈R;x≠2, x≠-1 , 2-e<y<27524751555750011810991210310(0,0)00(0,0)1285875305435y=2y=22667002334260x=-10x=-118383252458085x=2x=212287251905635y=e-200y=e-21152525138176000141890974326800222885048260007524755340351447800552450012096752482853295650276860The region R={x,y; -1<x<2, and 2-e<y<2}is the largest region for which the IVP has a unique solution.00The region R={x,y; -1<x<2, and 2-e<y<2}is the largest region for which the IVP has a unique solution.Question 2:Solve the initial value problemxy2+4xdx+8y-2x2ydy=0, x<2y0=0Solution:4895849583565Separable00Separable4429125609600xy2+4xdx=2yx2-4dyxx2-4dx= 2yy2+4dy122xx2-4dx= 2yy2+4dy12lnx2-4=lny2+4+cat 0,0: 12ln-4=ln4+cc= -12ln412lnx2-4=lny2+4-12ln4lny2+42=lnx2-4y2+42= x2-4∴ y2=4+2x2-4Question 2:Obtain the general solution of the differential equation2cosxdydx+ysinx-4x+52y3=0, x∈(-π2,π2)Solution:45339003775075right3632200Multiply by (cosx)0Multiply by (cosx)4152900295592544862752879725Linear00Lineardydx+12tanx y= 4x+522cosx y31y3 dydx+12tanx 1y2= 4x+522cosxLet u=y-2dudx= -2y-3dydxdydx= 1-2y-3dudx∴ 1y3 1-2y-3 dudx+12tanx u= 4x+522cosx dudx+-tanxu= 4x+52-cosxμx,y=e-tanx dx=elncosx=cosxcosx dudx+cosx-tanxu= 4x+52-cosx cosxPx= -sinx , Qx=4x+52ddxμ. u= μ.Qx → μ.Q(x)=μ.Qxdxcosx .u= -4x+52 dx = -1444x+52 dxcosx . u= -144x+533+c∴ cosxy2= -1124x+53+cQuestion 3:Solve the differential equationcosy+ycosxdx+(sinx-xsiny)dy=0Solution:Mx,y=cosy+y cosx , ?M?y=-siny+cosxNx,y=sinx-x siny , ?N?x=cosx-siny?M?y=?N?x → D.E is Exact.fx,y=Mx,ydx= cosy+y cosxdx=x cosy+y sinx+ φy?f?y=-x siny+sinx+ φ'y?f?y=Nx,y → -x siny+sinx+φ'y=sinx-x sinyφ'y=0φy=c∴fx,y → x cosy+y sinx=cQuestion 3:By using an appropriate substitution, or any other method, find the general solution of the differential equationdydx= yxlnxy, x>0, y>0Solution:Let u=xy →y= uxdydx= xdudx-ux2 xdudx-ux2=ux2lnuxdudx-u=ulnux dudx=ulnu+u1ulnu+u du= 1x dx1ulnu+u du= 1x dx1u[lnu+1] du= 1x dx1ulnu+1 du= 1x dxlnlnu+1=lnx+c∴D.E is →lnlnxy+1=lnx+c?Question 4: Let y(1+cosx)=C, where x∈(0,π), be a given family of curves. Find the family of orthogonal trajectories for this family of curves.Solution:46577251149985Separable00Separable42005251211580y -sinx+1+cosxdydx=0D.E of orthogonal trajectories → dydx= -1+cosxy sinxy sinx dy= -1+cosx dxy dy=-(1+cosx)sinx dxy dy= (-cscx-cotx) dxy22=lncscx+cotx-lnsinx+c ................
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