Question 1184644

The characteristic equation of the D.E. is  {{{lambda^2 + 4*lambda + 4 = 0}}}, which gives the double root {{{lambda = -2}}}.

===>  The general solution to the D.E. is {{{y = A*e^(-2t) + Bt*e^(-2t)}}}.


==> {{{"y'(1)" = (-2A+B)e^(-2) - 2Be^(-2) =1}}} ===> {{{(-2A-B)e^(-2) = 1}}} ===> {{{2A+B = -e^2}}}.


Also, {{{y(1) = Ae^(-2) + Be^(-2) = 0}}} ===> A + B = 0, or B = -A.


Combining this with the preceding equation, we get 


{{{2A - A = A = -e^2}}}.   ===> {{{B =e^2}}}   ===> {{{y = -e^2*e^(-2t) + e^2*t*e^(-2t)}}}.


after combining factors and simplifying, {{{y(t) = (t-1)e^(-2(t-1))}}}