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y''+2y'+y=2e^(-t) find the general solution of the giving nonhomogenous differential equation
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Lets first find the solution associated with the Homogeneous DE by solving the auxiliary equation \(m^2+2m+1=0 \implies (m+1)^2=0\). So we have a repeated root at \(m=-1\) and thus \(y_c=c_1e^{-t}+c_2te^{-t}\). is the solution associated with the HDE. Assume a particular solution \(y_p=At^2e^{-t}\) and substitute for it into the given DE to find A.
can you continue because i got confused when i did what you said
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