240-2008&2009-2-M20-May2009(1)

240-2008&2009-2-M20-May2009(1) - Kuwait University...

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Unformatted text preview: Kuwait University Math. 240 May 13, 2009 Dept. of Math. & Comp. Sc. Second Midterm Exam Duration 90 Minutes Calculators and mobile phones are not allowed Answer the following questions. 1. (3 Points) Let A = mD ‘ 2 and B : D + :L‘. Show that AB 74 BA. 2. (3 Points) Solve the initial value problem y"+4y : 0; y(0) = 3 and y'(0) : —4. 1 1 3. Let y] 2 xsin (—> and M 2 3:005 <—) . x m (a) (2 Points) Show that yl and yg are solutions of the equation $4y"+y=0; x>0. (b) (2 Points) Show that 3,11 and y2 are linearly independent functions on (0, 00) . 4. Use inverse operators to find the general solution of the following equations (a) (5 Points) (D3 — 3D2 + 4) y = 662I + 8x2 (b) (5 Points) y” + 2y’ + 2y : 28”” sinx + 10 cos Zr 5. (5 Points) Find the general solution of the equation nyll i $y1+ y 2 2:]: given that y1 = :c is a solution of the corresponding homogeneous equation. Some Useful Formulas 1 1 . we “W4 mm) 1 b _:z:sinb$ DMb‘zcos'm~~ 2b 1 g], bx__mcosb:c D2+b2‘“ ’ 2b 99 Math. 240 Second Midterm Solution May 13, 2009 1. A‘By = (xD—2)(Dy+xy) = x[D2y+ny+y] — 2(Dy+xy) => AB = xD2 + (x2 ~2)D—x. BAy = (D +x)(ny— 2y) = (xDzy + Dy — 2Dy) + 26(ny — 2y) :> BA = xDz + (x2 — l)D -— 2.x. 2. m2 +4 = 0 => m = i2i 2 y = c1c052x+czsin2x => y’ = —2c1 sin2x+2czc052x :> y = 3cost—25in2x. y(0)=3 :>c|=3 y’(0)=—4 2 C2 =—2 4- a- y :yC+yPl +Yp2 mE—smwo:<m+1><m—2>2=o: _ 1 ix _ 6x26” _ 2 2,:————6 7—- . y” (D+1)(D—2)2( e ) 3(2!) ” ypz = w~—D3_3102+4(8x2)=2(1+%DZ)(x2)= 4~ 13- y =yc+yp1+ypz m2+2m+2 = 0 :> (m+1)2+1= 0 :> ya = e'x(c1cosx+02sinx). ypl = —1—w(23"‘sinx) = DZe—X (sinx) = 2e“(lx%) = —xe"‘cosx. (D+1)2 +1 2 +1 _ 1 z 5 = _ = ' w 2 . y,,z — D2 +2D+ 2 (lOcost) D_1 (cost) (D+1)(cos2x) 251n2x cos x 5. y = y1u = xu, y' : xu' + u, y” = xu" + 221' u~+%u'=-x%7 xu’221nx+c1. u=(1nx)2+cmlntxl+cz y=x[cllan|+cz+(lnx)2]. . i \ 99 ...
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240-2008&amp;amp;2009-2-M20-May2009(1) - Kuwait University...

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