Question

Find the general (real) solution of the differential equation ( y' = dy/dx ):

y'' + 8y' + (73/4)y = 0; y(x) =
Find the unique solution that satisfies the initial conditions:
y(0) = 1 , y'(0) = -1; y(x) =

Answer 1

The general solution of the differential equation is y(x) = c1 * e^(-4x) * cos((7x)/2) + c2 * e^(-4x) * sin((7x)/2). The unique solution that satisfies the initial conditions is y(x) = 2 * e^(-4x) * cos((7x)/2) - e^(-4x) * sin((7x)/2).

Step-by-step explanation:

To find the general (real) solution of the differential equation, we first need to find the characteristic equation of the given equation.

The characteristic equation is given by r^2 + 8r + (73/4) = 0.

Solving this quadratic equation, we find two distinct real roots: r1 = -4 - (7i/2) and r2 = -4 + (7i/2). Therefore, the general solution of the differential equation is given by y(x) = c1 * e^(-4x) * cos((7x)/2) + c2 * e^(-4x) * sin((7x)/2), where c1 and c2 are arbitrary constants.

To find the unique solution that satisfies the initial conditions, we substitute the given values of y(0) = 1 and y'(0) = -1 into the general solution and solve for the constants c1 and c2.

By substituting the initial conditions, we get the unique solution: y(x) = 2 * e^(-4x) * cos((7x)/2) - e^(-4x) * sin((7x)/2).