Final answer:
The differential equation y' - y = 0 is a first-order linear homogeneous equation. The general solution is y = ±C'e^x, where C' is a constant. The solution provided by the student, y = e^x is a particular instance of the general solution.
Step-by-step explanation:
The differential equation given is a first-order linear homogeneous differential equation and can be solved using the method of separation of variables. The equation is y' - y = 0, where y' represents the derivative of y with respect to x.
To solve this equation, we can rearrange terms to express the derivative y' as y' = y. Now, separating the variables, we get dy/y = dx which allows us to integrate both sides. The integration of the left side with respect to y is ln|y|, and the integration of the right side with respect to x is x plus the constant of integration C.
After integrating, we have ln|y| = x + C. Exponentiating both sides gives us |y| = e^x * e^C. Since e^C is just another constant, we can denote e^C as C' (which could be positive or negative), we end up with y = ±C'e^x. The student provided y = e^x which is indeed a solution to the differential equation, but it represents the solution when the constant C' is exactly 1 and does not cover all possible solutions.