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Question

Consider the equation yy2y=0y”−y'−2y=0. (a) Show that y1(t)=ety_1(t)=e^{−t} and y2(t)=e2ty_2(t)=e^{2t} form a fundamental set of solutions. (b) Let y3(t)=2e2ty_3(t)=−2e^{2t}, y4(t)=y1(t)+2y2(t)y_4(t)=y_1(t)+2_{y2}(t), and y5(t)=2y1(t)2y3(t)y_5(t)=2_{y1}(t)−2_{y3}(t). Are y3(t)y_3(t), y4(t)y_4(t), and y5(t)y_5(t) also solutions of the given differential equation? (c) Determine whether each 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)][y_1(t),y_3(t)];[y_2(t),y_3(t)];[y_1(t),y_4(t)];[y_4(t),y_5(t)].

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We were given

yy2y=0(1)y''-y'-2y=0 \tag{1}

and y1(t)=ety_1(t)=e^{-t} and y2(t)=e2ty_2(t)=e^{2t}.

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