Final answer:
The general solution of the second-order differential equation 2y'' - 5y' + 6y = 0 involves finding the roots of the characteristic equation. These roots dictate the form of the solution, which often includes exponential and trigonometric functions. The student must verify their specific solution by substituting it and its derivatives back into the original equation.
Step-by-step explanation:
To find the general solution of the second-order differential equation 2y'' - 5y' + 6y = 0, we look for solutions of the form y(x) = e^(rx), where r is a constant that will be determined. Substituting y and its derivatives into the equation gives us the characteristic equation 2r^2 - 5r + 6 = 0. Solving this quadratic equation for r gives the roots necessary to construct the general solution.
Since these roots are typically real and distinct, or complex conjugates, the general solution is often a combination of exponential and trigonometric functions, which may look like Ak cos(kx) + Bk sin(kx), where Ak and Bk are constants and k is related to the imaginary part of the complex roots, if any.
The student provided a more specific form of the solution, which involves complex numbers: y(x) = e^(45x)(C1 cos(423x) + C2 sin(423x)). To prove that this is, in fact, the solution to the original differential equation, one needs to take the derivatives and substitute them in the differential equation to see if it is satisfied.