Question

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ \begin{array}{|l|l|l|l|} \hline \mathrm{C} / \mathrm{G} & y^{\prime \prime}+y^{\prime}-2 y=-10 e^{-t}+5 \delta(t-1), & y(0)=7, & y^{\prime}(0)=-9 \end{array} $$

Short answer

Question: Given the second-order linear inhomogeneous differential equation \(y''(t) + y'(t) - 2y(t) = -10e^{-t} + 5\delta(t-1)\) with initial conditions \(y(0) = 7\) and \(y'(0) = -9\), obtain the general solution and illustrate the solution graphically. Solution: The general solution to the initial value problem is given by \[y(t) = y_c(t) + y_p(t) = (C_1 e^t + C_2 e^{-2t}) - 10te^{-t} - 5(t-1)e^{-(t-1)}u(t-1).\] To find the value of the constants \(C_1\) and \(C_2\), we substitute \(t = 0\) into the equation and compare with the values from the initial conditions. The graphical representation of the solution will include an exponential function that shifts at \(t = 1\). Refer to the function plot from your computer software for a visual representation.

Step-by-step solution

Solve the homogeneous equation

To find the complementary function, we first solve the homogeneous version of the given differential equation: $$y''(t) + y'(t) - 2y(t) = 0.$$ This is a second-order linear homogeneous equation with constant coefficients, which has a characteristic equation given by: $$r^2 + r - 2 = 0.$$ To find its roots, we can either use the quadratic equation formula or factor it. In this case, it factors easily: $$(r - 1)(r + 2) = 0,$$ yielding two roots: \(r_1 = 1\) and \(r_2 = -2\). Thus, the complementary function is given by: $$y_c(t) = C_1 e^t + C_2 e^{-2t}.$$

Find the particular solution

To find the particular solution, \(y_p(t)\), we must address the non-homogeneous part of the equation, namely \(-10e^{-t} + 5\delta(t-1)\). However, the presence of the Dirac delta function will require us to split the particular solution into two parts. For the continuous part, \(-10e^{-t}\), we can make the initial guess that \(y_p(t) = Ate^{-t}\), as it is of the same form as the forcing term (with \(A\) being a constant). Taking the first and second derivatives of the guess yields: $$y_p'(t) = A(-te^{-t} + e^{-t}), \hspace{1cm} y_p''(t) = A(t e^{-t} - 2e^{-t}).$$ Substituting these into the main equation gives: $$A(t e^{-t} - 2e^{-t}) + A(-te^{-t} + e^{-t}) - 2(Ate^{-t}) = -10e^{-t}.$$ Simplifying and comparing the coefficients, we find that \(A = -10\), so the particular solution for this part is: $$y_{p1}(t) = -10te^{-t}.$$ For the part involving the delta function, we start by considering the Laplace transform of the equation. We can deal with the delta function more straightforwardly in the Laplace domain. Let the Laplace transform of \(y(t)\) be \(Y(s)\), applying the Laplace transform to both sides of the given equation, we get $$s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-2Y(s) = -\frac{10}{s+1}+5e^{-s}.$$ Substituting the initial conditions and solving for \(Y(s)\), we have $$Y(s) = \frac{2s^3+14s^2+34s+10}{s(s+1)(s+2)}.$$ Taking the inverse Laplace transform of this gives us the particular solution for the delta function part, \(y_{p2}(t)\): $$y_{p2}(t) = -5(t-1)e^{-(t-1)}u(t-1),$$ where \(u(t-1)\) is the unit step function, used to "turn on" the response only after \(t = 1\). Thus, the overall particular solution \(y_p(t)\) is the sum of the two parts: $$y_p(t) = -10te^{-t} - 5(t-1)e^{-(t-1)}u(t-1).$$

Apply initial conditions

Since we have already found the complementary function with unknown constants, there is no need to apply the initial conditions to the particular solution. In this case, the initial conditions are only used to determine the constants in the homogeneous solution.

Form the general solution

The general solution to the initial value problem is the sum of the complementary function and the particular solution: $$y(t) = y_c(t) + y_p(t) = (C_1 e^t + C_2 e^{-2t}) - 10te^{-t} - 5(t-1)e^{-(t-1)}u(t-1).$$ This is the solution of the given initial value problem, and the graph of this solution will involve an exponential function that will shift at \(t = 1\).

Key concepts

Homogeneous Differential Equation

In solving initial value problems, understanding homogeneous differential equations is crucial. A homogeneous differential equation is one where there is a distinct relationship between a function and its derivatives without any external forcing term, or "input," in the equation. In such equations, the right side is zero. For example, the given homogeneous equation \[ y''(t) + y'(t) - 2y(t) = 0 \] signifies an equation where the external force does not influence its behavior.

To solve this, we generally convert the differential equation into a characteristic equation. Here, the characteristic equation is obtained by substituting the exponential function into the differential equation, leading to \[ r^2 + r - 2 = 0 \]. By solving the characteristic equation using factoring or the quadratic formula, we find the roots, which in this case are \( r_1 = 1 \) and \( r_2 = -2 \).

The solution to the homogeneous differential equation involves finding the complementary function, which includes these roots.

Complementary Function

The complementary function is a solution to the homogeneous differential equation. This function represents the natural behavior of the system when no external inputs are present. In the homogeneous case, the final solution is formed with the exponential terms derived from the roots of the characteristic equation.

For instance, with the roots \( r_1 = 1 \) and \( r_2 = -2 \), the complementary function becomes: \[ y_c(t) = C_1 e^{t} + C_2 e^{-2t} \]. Here, \( C_1 \) and \( C_2 \) are constants that can be determined by the initial conditions. These constants adjust the weight of each exponential component in the final solution.

The complementary function captures the baseline dynamic response of the system and acts as a foundation for constructing the overall solution when combined with a particular solution.

Particular Solution

To fully solve a non-homogeneous differential equation, we must find a particular solution that accounts for the non-homogeneous part of the equation. The particular solution specifically addresses the external forcing terms, like constants or functions that cause the system to be driven or influenced.

In this exercise, we have non-homogeneous components \(-10e^{-t} + 5\delta(t-1)\). This requires finding solutions that individually address each component. For the \(-10e^{-t}\) term, we assume a trial solution that matches its form which was \(-10te^{-t}\). Through substitution into the differential equation, and equating coefficients, we determine the values of constants involved.

The delta function component \(5\delta(t-1)\) introduces an impulse at \(t = 1\), parsed using the Laplace transform for managing instantaneous changes at specific time points.

When both parts are solved, they combine to give the full particular solution.

Laplace Transform

The Laplace Transform is a powerful mathematical tool utilized to convert complex differential equations from the time domain into a simpler algebraic form in the frequency domain. This transformation is especially useful for solving equations with inputs like the Dirac delta function, as seen in our non-homogeneous term \(5\delta(t-1)\).

By applying the Laplace Transform on the original equation, it facilitates handling complex terms more straightforwardly. The Dirac delta function, which is pivotal in engineering and physics, acts as an idealized "impulse" or "kick." The transform of a delta function prompts a pivotal change at the specific time where it "acts." The Laplace of \( y''(t) + y'(t) - 2y(t) = -10e^{-t} + 5\delta(t-1) \) transforms it into an equation in the \( s \)-domain.

The zeros resulting from transformation are solved back into the time domain using inverse Laplace techniques. Thus, allowing us to smoothly combine this particular solution back with the complementary function for the complete solution.