Question

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

Answer 1

`y'''-2y''+y' = 0`

Explanation:

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.