y′′(t)+2y′(t)−3y(t)=e−3t,y′(0)=0,y(0)=0y(t)→Y(p)p2Y(p)−py(0)−y′(0)+2(pY(p)−y(0))−3Y(p)=p+31(p2+2p−3)Y(p)=p+31Y(p)=(p+3)2(p−1)1(p+3)2(p−1)1=(p+3)2A+p+3B+p−1C1=A(p−1)+B(p+3)(p−1)+C(p+3)2(B+C)p2+(A+2B+6C)p+(−A−3B+9C)=1⎩⎨⎧B+C=0A+2B+6C=0−A−3B+9C=1⇒A=−41,B=−161,C=161Y(p)=−4(p+3)21−16(p+3)1+16(p−1)1↦−41te−3t−161e−3t+161ety(t)=−41te−3t−161e−3t+161et
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