Firstly, I haven't seen the notation I used in alevel but I just used it for the sake of ease of typing it online.1st. Sub in the trial solution f(x)= Ae^(mx) and its derivatives- f'(x)= Ame^(mx) and f''(x)= Am^(2)e^(mx). Simplify by dividing by Ae^(mx) to get m^2+ 3m + 2= 0.Solve the quadratic by inspection to the solutions m=1 and m=2. Since when each solution is substituted into the original differential the result =0 we can say that the sum of the solutions is correct. (0+0=0). So the solution is f(x)= Ae^x +Be^2x
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