Final answer:
The differential equation y''-2y'+2y=0 has a general solution expressed as e^x(A cos(x) + B sin(x)), which fits the specific form y(x) = Ce^x sin(x) with appropriate substitutions.
Step-by-step explanation:
We are tasked with solving the second order linear homogeneous differential equation y''-2y'+2y=0. This can be solved using the characteristic polynomial, which is obtained by replacing y with r, y' with r^2, and y'' with r^3 to get the quadratic equation r^2 - 2r + 2 = 0.
Using the quadratic formula, the roots of the characteristic polynomial are r = 1 ± i. Since these roots are complex, we express the general solution in terms of sine and cosine functions multiplied by an exponential function. Therefore, the general solution to the differential equation in question is y(x) = e^x(A cos(x) + B sin(x)).
However, we're asked to express the solution in the specific form ce^(ax) sin(βx + γ). We can match this form by setting A = 0 and B = C, and observing that β = 1 and γ = 0. Therefore, the solution can be rewritten as y(x) = Ce^x sin(x).