Why do you think, your solution is wrong? It's perfectly right! It's the general solution of the equation (I've not checked it, but this you can do by taking the derivative and check whether it fulfills your equation), and whether you call the integration constant [itex]-C_3[/itex] or [itex]C[/itex] doesn't matter.
The solution is made unique by giving, e.g., an initial value [itex]y(0)=y_0[/itex] as a constraint. If you use this in your and the textbook's answer you'll get the same result by choosing the right [itex]C_3[/itex] and [itex]C[/itex] to match the initial-value constraint, respectively.