We want to find a differential equation whose general solution is y=c_(1)e^(-2t)+c_(2)te^(-2t)...(1) We know htat y=c_(1)e^(lambda_(1)t)+c_(2)te^(lambda_(1)t) is general solution of the equation ay''+by' + cy=0 iff lambda_(1) is the only root of the quadratic equation alambda^(2)+blambda + c=0...(2) Therefore, we need to find coefficients a,b and c such that lambda_(1)=lambda_(2)=-2 is the solution of the Eq. (2). The quadratic equation whose only root is -2 is (lambda +2)^(2)=lambda^(2)+4lambda + 4 Therefore, one of the differential equations whose general solution is Eq. (1) is y''+4y'+4=0 Result: One example is y''+4y'+4=0.