What is the differential equation of which #y = (c_1+c_2x)e^x+c_3# is the general solution?

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Answer 1 · 2018-02-24T22:03:32

#y'''-2y''+y' = 0#

Given general solution:

#y = (c_1+c_2x)e^x+c_3#

First note that due to the arbitrary constant #c_3# the differential equation will not involve #y# directly - only its derivatives.

Let us write the first few derivatives:

#y' = (c_1+c_2+c_2x)e^x#

#y'' = (c_1+2c_2+c_2x)e^x#

#y''' = (c_1+3c_2+c_2x)e^x#

We can see that these are in arithmetic progression and hence:

#y'''-2y''+y' = 0#

Note that we have eliminated #3# constants in taking the derivative #3# times, so this is the required differential equation.