Question

Solve the given initial value problem: y′′−2y′+2y=cost

Answer 1

To solve the given initial value problem, we need to find the particular solution for the differential equation y'' - 2y' + 2y = cost. Plugging this form into the differential equation will allow us to solve for the constants A and B.

Step-by-step explanation:

To solve the given initial value problem, we need to find the particular solution for the differential equation. The general solution is given by y(t) = c1e^(t)cos(t) + c2e^(t)sin(t), where c1 and c2 are constants. To find the particular solution, we need to consider the form of the right-hand side, which is cost. Since the right-hand side is a cosine function, we can assume a particular solution of the form yp(t) = Acos(t) + Bsin(t), where A and B are constants. Plugging this into the differential equation and solving for A and B will give us the particular solution.

Answer 2

The solution to the given initial value problem
`\(y'' - 2y' + 2y = \cos(t)\) is \(y(t) = e^t(A\cos(t) + B\sin(t)) + (1)/(3)\cos(t)\`), where (A) and (B) are constants determined by the initial conditions.

Step-by-step explanation:

To solve the given second-order linear homogeneous differential equation, we first find the characteristic equation. The characteristic equation for the given differential equation (y'' - 2y' + 2y = 0) is
`\(r^2 - 2r + 2 = 0\)`. Solving this quadratic equation gives the characteristic roots (r = 1 pm i). The general solution for the homogeneous part is
`\(y_h(t) = e^t(A\cos(t) + B\sin(t))\), where \(A\) and \(B\)` are constants.

Now, to find the particular solution for the non-homogeneous part
`\(y_p(t) = (1)/(3)\cos(t)\)`, we use the method of undetermined coefficients. Since the right-hand side is (cos(t)), we assume a particular solution of the form
`\(y_p(t) = C\cos(t) + D\sin(t)\)`. Plugging this into the original differential equation and solving for the coefficients (C) and (D) yields
`\(C = (1)/(3)\) and \(D = 0\)`.

Therefore, the complete solution is the sum of the homogeneous and particular solutions:
`\(y(t) = y_h(t) + y_p(t) = e^t(A\cos(t) + B\sin(t)) + (1)/(3)\cos(t)\)`. The constants (A) and (B) can be determined using the initial conditions provided for the specific problem.