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D² + 3D + 2) y = e^(e^x) can anybody provide me the full solution of this problem

Romil

on 2016-03-06 10:30:00  

solve for complementary solution ( D² + 3D + 2) y =0 m^2+3m+2=0 (m+2)(m+1)=0 m=-2,m=1 yc=c1exp(-2t)+c2exp(t) Solve for yp D(yc)=-2c1exp(-2t)+c2exp(t) D^2(yc)=4c1exp(-2t)+c2exp(t) substitute the derivative of yc to the given equation -2c1exp(-2t)+c2exp(t)+3[4c1exp(-2t)+c2... 2[c1exp(-2t)+c2exp(t)]=e^(e^t)