91Ó°ÊÓ

The complementary solution is found by solving the homogeneous part of a differential equation. For the given exercise, the homogeneous form is: \[ y''' - y'' = 0 \]This implies solving the equation without considering the non-homogeneous part (the term with \(t^2\)). We find the characteristic equation, which helps determine the roots. Here, we have:\[ r^3 - r^2 = 0 \]This can be factored as:\[ r^2 (r - 1) = 0\]From this, we get the roots \(r = 0, 0, 1\). The roots help form the complementary solution, which is:\[ y_c(t) = C_1 + C_2 t + C_3 e^t \]This solution represents the general solution to the homogeneous equation. The constants \(C_1, C_2, C_3\) will later be determined using initial conditions.

The particular solution tackles the non-homogeneous part of the differential equation. In the exercise, the term \(3t^2\) requires us to guess a form for \(y_p(t)\). Given the polynomial on the right, we try a polynomial solution. Here, we assume:\[ y_p = At^4\]We then differentiate: \[ y_p''' = 24At \] and \[ y_p'' = 12At \]Substituting these into the original equation:\[ 24At - 12At = 3t^2 \]Solving, we get:\[ 12At = 3t^2 \]Thus, \(A = \frac{1}{4}\). So, the particular solution is:\[ y_p(t) = \frac{1}{4}t^4 \]. This accounts for the non-homogeneous term in the differential equation.

Initial conditions help specify the values of the constants \(C_1, C_2, C_3\) in the complementary solution. The initial conditions given are:\[ y(0) = 0 \] , \[ y'(0) = 0 \] , \[ y''(0) = 0 \]We start by combining the complementary and particular solutions:\[ y(t) = C_1 + C_2 t + C_3 e^t + \frac{1}{4}t^4 \]Evaluating at \(t = 0\):\[ y(0) = C_1 = 0 \]By differentiating, we find:\[ y'(t)= C_2 + C_3 e^t + t^3 \]Again at \(t = 0\):\[ y'(0) = C_2 = 0 \]And finally, the second derivative:\[ y''(t) = C_3 e^t + 3 t^2 \]For \(y''(0)\):\[ y''(0) = C_3 = 0 \]Thus, the constants \(C_1, C_2, C_3 \) all equal zero.

The characteristic equation comes from the homogeneous part of a linear differential equation with constant coefficients. For the exercise, it originates from:\[ y''' - y'' = 0 \]Substitute \(y = e^{rt}\), resulting in:\[ r^3 e^{rt} - r^2 e^{rt} = 0 \]Simplifying to:\[ r^3 - r^2 = 0 \]Factoring gives:\[ r^2 (r - 1) = 0 \]This provides the roots \(r = 0, 0, 1 \), crucial for forming the complementary solution:\[ y_c(t) = C_1 + C_2 t + C_3 e^t \]. The characteristic equation directly relates to the types of solutions we formulate for the homogeneous differential equation.

The general solution of a non-homogeneous differential equation is the sum of the complementary and particular solutions. For the given exercise, the complementary solution is:\[ y_c(t) = C_1 + C_2 t + C_3 e^t \]The particular solution is:\[ y_p(t) = \frac{1}{4}t^4 \]Combining these gives us the general solution:\[ y(t) = y_c(t) + y_p(t) \]Thus, the general solution is:\[ y(t) = C_1 + C_2 t + C_3 e^t + \frac{1}{4}t^4 \]Using initial conditions, we find the constants all to be zero. So the simplified general solution becomes:\[ y(t) = \frac{1}{4}t^4 \]. This solution represents the function that satisfies the original differential equation under given initial conditions.