Ok, so 2nd derivative is .5 times first derivative, means if that was your differential equation, then y=C1*e^(.5*x) is one fundamental solution, and y=C2 is the other. At x=0, there are the boundary conditions y=0 and y'=3. You stated it really badly by the way. The first few times I read it, I thought you meant d^2y/dx^2 = 0.5dy/dx was true only at y=0, and dy/dx=3 is true when x=0, which would be 2 boundary conditions and no differential equation. So obviously C1=-C2 so that y(0)=0. Then y'=.5*C1=3 at x=0, so C1=6. y=6*(e^(.5*x)-1). If x=5, then y=6*(e^(2.5)-1). Apparently you didn't go wrong, since you got the same answer I did, although I never involved the natural log.