The differential equation is and the function that is to be proven as solution is

y = ae^2x + be^–x, now we need to find the value of and .

= 2ae^2x – be^–x

= 4ae^2x + be^–x

Putting these values in the equation, we get,

4ae^2x + be^–x –(2ae^2x – be^–x) – 2(ae^2x + be^–x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.