Consider the equation y" - y' - 2y = 0. Assume that y₁ (t) (b) Let Y3 (t) Y4 (t) = et and y₂(t) = 2e²t = y₁ (t) + 3y₂ (1

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Consider The Equation Y Y 2y 0 Assume That Y T B Let Y3 T Y4 T Et And Y T 2e T Y T 3y 1 1 (257.36 KiB) Viewed 14 times

Consider The Equation Y Y 2y 0 Assume That Y T B Let Y3 T Y4 T Et And Y T 2e T Y T 3y 1 2 (117.34 KiB) Viewed 14 times

Consider The Equation Y Y 2y 0 Assume That Y T B Let Y3 T Y4 T Et And Y T 2e T Y T 3y 1 3 (193.27 KiB) Viewed 14 times

Consider the equation y" - y' - 2y = 0. Assume that y₁ (t) (b) Let Y3 (t) Y4 (t) = et and y₂(t) = 2e²t = y₁ (t) + 3y₂ (1) - ys(t) = 2y₁ (t) — 3 y3(t). e2t form a fundamental set of solutions. Are y3 (1), y4 (t), and y5 (t) also solutions of the given differential equation? O Yes O No O Impossible to tell (c) Determine which of the following pairs forms a fundamental set of solutions: [y₁ (t), y3(t)]; [y2(t), y3(t)]; [y₁(t), y4(t)]; [y4(t), y5(t)]. ○ [y₁ (t), y3(t)], [y2(t), y3(1)], and [y4(t), ys(t)] ° [y₁ (t), y3(t)], [y₁(t), y4(t)], and [y4(t), y5(t)] O [2(1), 3(1)] and [yi(t), y4 (t)], and [y4 (t), ys(t)] O [y4(1), ys (1)] and [y₂(1), 3(1)] ○ [y₁ (t), y3 (t)] and [y₁(t), y4(t)]
Question 5 of 17 < View Policies Current Attempt in Progress Find the general solution of the differential equation y" + 2y + 5y = 2 sin(2t). NOTE: Use c₁ and ce for the constants of integration. y(t) = =
Question 6 of 17 View Policies Current Attempt in Progress y = Use the method of undetermined coefficients to solve the initial value problem. y = -1 8 < -cos2t - 40 cos2r 43 40 -cos2t -1 10 43 sin2t + -1 cos2t + sin2t + 10 -cos2t 43 40 > 43 40 43 40 -sin2t + -sin2t + -1 -cos2t + -sin2t 10 3 4 4 1 8 3 4 1 2 8 4 4 + y" + 4y = 1² +6e¹, y(0) = 0, y'(0) = 1 1 + + 2 6 1 + + بريطان به نام مال 1 6 Ser
Question 7 of 17 View Policies Current Attempt in Progress < Find the solution of the initial value problem 1 13y" - 14y + y = 0, y(0) = 13, y'(0) = 13 Then determine the maximum value M of the solution and also find the point where the solution is zero. y(t) = M = The point where the solution is zero is