1 votes 1 votes The solution of differential equation $y''+3y'+2y=0$ is of the form $C_1e^x+C_2e^{2x}$ $C_1e^{-x}+C_2e^{3x}$ $C_1e^{-x}+C_2e^{-2x}$ $C_1e^{-2x}+C_2e^{-x}$ Calculus gate1995 calculus out-of-gate-syllabus + – Kathleen asked Oct 8, 2014 • recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 1.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 0 votes 0 votes ans is c. here y''-(-3)y'+2y=0 therefore y=c1e^2x + c2e^x and solving we get c1e^-x + c2e^-2x jayendra answered Dec 27, 2014 • selected Jun 2, 2015 by Rajarshi Sarkar jayendra comment Share Follow See 1 comment See all 1 1 comment reply anchitjindal07 commented Dec 19, 2018 reply Follow Share What is difference between option C and option D 0 votes 0 votes Please log in or register to add a comment.