91Ó°ÊÓ

In second-order differential equations, we often come across the term 'homogeneous.' A homogeneous equation is one without an extra term involving the independent variable (in this case, t). For example, in the equation provided: \[ y'' - 4y = 32t \] the equivalent homogeneous equation would be: \[ y'' - 4y = 0 \] To solve this, we use the characteristic equation, which helps us determine the type of exponential functions that satisfy our differential equation. In this example:\[ r^2 - 4 = 0 \] Solving for r, we get the roots: \[ r = \pm 2 \] Therefore, the general solution to the homogeneous part is \[ y_h(t) = C_1 e^{2t} + C_2 e^{-2t} \] where C_1 and C_2 are constants to be determined.

For non-homogeneous equations, we need to find a particular solution that deals with the non-zero term. In our example: \[ y'' - 4y = 32t \] The non-homogeneous term here is 32t. We assume a particular solution in a similar form, which is \[ y_p(t) = At + B \] Differentiating, we get:\[ y_p'(t) = A \] \[ y_p''(t) = 0 \] Substituting these into the original equation, we obtain: \[ 0 - 4(At + B) = 32t \] This simplifies to: \[ -4At -4B = 32t \] By matching coefficients, we get two equations: \[ -4A = 32 \Rightarrow A = -8 \] \[ -4B = 0 \Rightarrow B = 0 \] Thus, our particular solution is: \[ y_p(t) = -8t \].

Once we have the general solution, which is the sum of the homogeneous and particular solutions: \[ y(t) = C_1 e^{2t} + C_2 e^{-2t} - 8t \] we can apply initial conditions to find the specific solution for our problem. The initial conditions given are: \[ y(0) = 0, \quad y'(0) = 6 \] Substitute t = 0 into the general solution: \[ 0 = C_1 + C_2 \Rightarrow C_2 = -C_1 \] Next, use the condition for the first derivative of y. Calculate the derivative: \[ y'(t) = 2C_1 e^{2t} - 2C_2 e^{-2t} - 8 \] Substitute t = 0 and use y'(0) = 6: \[ 6 = 2C_1 - 2(-C_1) - 8 \] Simplifying, we get: \[ 6 = 4C_1 - 8 \Rightarrow 4C_1 = 14 \Rightarrow C_1 = 3.5 \] Then, \[ C_2 = -3.5. \]

The characteristic equation is a vital tool in solving homogeneous linear differential equations. It is derived from the differential equation by substituting a trial solution of the form \[ y = e^{rt}. \] For the homogeneous equation \[ y'' - 4y = 0, \] substitute \[ y = e^{rt}, \] and differentiate: \[ y'' = r^2 e^{rt}. \] Replacing into the original differential equation: \[ r^2 e^{rt} - 4 e^{rt} = 0 \] Factoring out the exponential term: \[ e^{rt} (r^2 - 4) = 0. \] Since the exponential function is never zero, solve: \[ r^2 - 4 = 0 \]. This yields the characteristic equation: \[ r^2 - 4 = 0. \] Solving the characteristic equation gives the roots: \[ r = \pm 2. \] These roots help us form the general solution of the homogeneous equation: \[ y_h(t) = C_1 e^{2t} + C_2 e^{-2t}. \]