Question

Find the general solution to 2y′′+50y=0. Give your answer as y=…. In your answer, use c_1 and c_2 to denote arbitrary constants and x the independent variable. Enter c _1 as c_1 and c_2 as c_2. help (equations)

Answer 1

To find the general solution to the differential equation 2y′′+50y=0, we assume a solution of the form y = e^(kx), where k is a constant. By solving the characteristic equation, we find two complex roots. The general solution is given by y = c_1cos(5√2x) + c_2sin(5√2x), where c_1 and c_2 are arbitrary constants.

Step-by-step explanation:

The given equation is 2y′′+50y=0. To find the general solution, we assume a solution of the form y = e^(kx), where k is a constant. Plugging this into the equation, we get k^2e^(kx) + 50e^(kx) = 0. Dividing both sides by e^(kx), we have the characteristic equation k^2 + 50 = 0.

Solving this equation, we find two solutions: k = ±sqrt(-50). Since the square root of -50 is a complex number, we can write it in terms of the imaginary unit i as k = ±5i√2.

Therefore, the general solution to the given differential equation is y = c_1cos(5√2x) + c_2sin(5√2x), where c_1 and c_2 are arbitrary constants.