Question

Find the solution of the given initial value problem: y"+y' = sec(t), y(0) = 6, y'(0) = 3, y'(0) = −4. y(t) = 2+4 cos(t) + 4 sin(t) — t cos(t) + sin(t) In(cos(t)) X

Answer 1

To solve the initial value problem y'' + y' = sec(t), with initial conditions y(0) = 6 and y'(0) = 3, use the method of variation of parameters. The solution is y(t) = 2 + 4cos(t) + 4sin(t) - tcos(t) + sin(t) * ln(cos(t)).

Step-by-step explanation:

The given initial value problem is y'' + y' = sec(t), with the initial conditions y(0) = 6 and y'(0) = 3. To solve this, we can use the method of variation of parameters. Let's assume the solution is of the form y(t) = u(t)cos(t) + v(t)sin(t), where u(t) and v(t) are unknown functions. Substituting this into the differential equation, we can solve for u'(t) and v'(t). After finding u(t) and v(t), we can substitute these back into the assumed solution to get the final answer.

In this case, the solution to the initial value problem is y(t) = 2 + 4cos(t) + 4sin(t) - tcos(t) + sin(t) * ln(cos(t)).

Answer 2

To find the solution of the given initial value problem, we need to solve the homogeneous equation and find the particular solution of the non-homogeneous equation using variation of parameters. The general solution is y(t) = A*e^(-t) + B - cos(t)ln|cos(t)|.

Step-by-step explanation:

The given initial value problem is: y'' + y' = sec(t), with the initial conditions y(0) = 6 and y'(0) = 3. To solve this, we need to find the general solution of the homogeneous equation y'' + y' = 0, which is y(t) = A*e^(-t) + B. For the particular solution of the non-homogeneous equation y'' + y' = sec(t), we can use variation of parameters. We find that yp(t) = -cos(t)ln|cos(t)|, so the general solution to the initial value problem is y(t) = A*e^(-t) + B - cos(t)ln|cos(t)|.