SOLVE Q22 Q10) The general solution of y" +2y" -y'-2y = 0, is:a) y(x) = c₂e²x+c₂e-² +₂exc) y(x) = c₂e²x + c₂e*b) y(x) = c₂e-2x + c₂exQ11) L(y") =d) y(x) = c₂e-²x + c₂e²+G₂e™*a) s² L(y(x)) + sy(0) - y'(0)c) s² L(y(x)) -sy(0)-y'(0)Q12) The kernel of the Laplace transformb) e-tb) s² L(y(x))+ sy(0) + y'(0)of f(t),t> 0, is:d) s² L(y(x))-sy(0) + y'(0)c) estQ13) Evaluate (e-t sin4t):a)842b)342x² +8+20d) +20Q14) If the power series method was used to solve the following ODE.(x-0.5)y"x+0.5y' + (x²-1) y = 0.xo = 0. Then the interval of convergence, is:b) (-1,1)c) (0,00)(9)Q15) Assume that the roots of a characteristic polynomial of a homogeneous ODE with constant coefficients are0,0,0,2+51,2-51, then the general solutiom of this ODE, is:a) c₂ + ₂x + ₂x² + e2x [Acos(x) + Bsin(x)]b) c₂ + ₂x + ₂x² +e5x (Acos(2x) + Bsin(2x)]c) c₂x + ₂x² + 3x³ + e*[Acos(5x) + Bsin(5x)]d) G₂ + G₂x + C₂x² + ²x [Acos(5x) + Bsin(5x)]Q16) Combine the following power series expressions into a single power series.Σ(n+1)(x-2)-1 +Eon(x - 2)"a) (2n + 1)(x)"b) To(2n + 2)(x)"c) Eno(2n + 2)(x - 2)"d)(2n+1)(x-2)"Q17) If you know that the radius of convergent of the series method for the ODEy"+y+=0, x=2 is 5. Find the value for b, (b<0):(a)-5(b)-3(c)-2-8Q18) The formula of the particular solution yp of y) + 4y" = 3 sin(2t) - 5cos2t, is:a) yp= Asin(t) + Bcos(t)b) ypAtsin(t) + Btcos(t).c) y₂ = Asin(2t) + Bcos(2t)d) y = Atsin(2t) + Btcos(2t)Q19) est dt =a)b) 09)d) ∞oQ20) The singular point (s) of (x + 1) y' + x²y = 0, is (are):a) 0,1b) 0,-1c) 0d)-11=b) 1-cosht c) (cosh2t - 1) d) (1-cosh2t)2t2y" +ty' - 3y = 0,t> 0, then y₂ (t) is:Q21) L1a) cosht - 1s(s²-1)Q22) Given that y₁ (t) = t¹ is solution fora)t b) tiQ23) L-1c) tid) taestsinst-atsinst1a)b)est sin3t5c)e-at sin3t5d)55(s-3)2+25AQ24) The general solution for y' = 6y²x, is:a) == 3x² + cb) ² = x² + cc) =² = 3x² + cd) == x² + cyQ25) The form of a particular solution of y"-4y' - 12y = sin(2t), is:a) y(t) = A sin(2t)b) y(t) = A cos(t)c) y(t) = A cos(2t) + B sin(2t)d) y(t) = A cos (t) + B sin(t)

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Q20) The singular point (s) of (x + 1) y' + x²y = 0, is (are): a) 0,1 b) 0,-1 c) 0 d)-1 021) - (1) a) cosht-1 b) 1-cosht c) (cosh2t - 1) 2t2y" + ty' - 3y = 0,t> 0, then y₂ (t) is: Q22) Given that y₁ (t) = t¹ is solution for a)t b) ti 1 1 e-3t sinst Q23) a) b) est sin3t 5 c) e-at sin3t 5 d) (S-3)2+25A 5 Q24) The general solution for y' = 6y²x, is: a)- = 3x² + c b) == x² + c c) == 3x² + c d) =/² = x² + c y Q25) The form of a particular solution of y"-4y' - 12y = sin(2t), is: b)u(t) A cos(t) d) (1-cosh2t) estsinst 5