Question

In Exercises 20-23 solve the initial value problem and plot the solution. $$ y^{\prime \prime}-3 y^{\prime}-10 y=7 e^{-2 x}, \quad y(0)=1, \quad y^{\prime}(0)=-17 $$

Short answer

Question: Solve the initial value problem given by the following second-order inhomogeneous differential equation with initial conditions: \(y'' - 3y' - 10y = 7e^{-2x}\), \(y(0) = 1\), and \(y'(0) = -17\).

Step-by-step solution

Find the complementary function

To find the complementary function, we need to solve the homogeneous equation: $$ y'' - 3y' - 10y = 0 $$ We can solve this by finding a function y in the form of \(y(x) = e^{mx}\), and then solving the characteristic equation: $$ m^2 - 3m - 10 = 0 $$ By factoring the equation, we get \((m - 5)(m + 2) = 0\). Thus, the complementary function is: $$ y_c(x) = C_1e^{5x} + C_2e^{-2x} $$

Find a particular solution

Now, we need to find a particular solution, y_p, which satisfies the given inhomogeneous equation: $$ y'' - 3y' - 10y = 7e^{-2x} $$ Since the right side of the equation contains a term of the form \(e^{-2x}\), we can assume a particular solution of the form: $$ y_p(x) = A x e^{-2x} $$ Now we find the first and second derivatives of \(y_p(x)\): $$ y_p'(x) = Ae^{-2x}(1 - 2x) \\ y_p''(x) = 2Ae^{-2x}(2x - 1) $$ Next, we substitute these derivatives into the inhomogeneous equation: $$ 2Ae^{-2x}(2x - 1) - 3Ae^{-2x}(1 - 2x) - 10A x e^{-2x} = 7e^{-2x} $$ We can cancel out the \(e^{-2x}\) terms: $$ 2A(2x - 1) - 3A(1 - 2x) - 10A x = 7 $$ Simplify the left side equation to solve for A: $$ A(13x - 7) = 7 $$ Divide by \((13x - 7)\), and we find the particular solution: $$ y_p(x) = \frac{7}{13x - 7} xe^{-2x} $$

Find the general solution

Now we can combine the complementary function and particular solution to find the general solution of the inhomogeneous equation: $$ y(x) = y_c(x) + y_p(x) = C_1e^{5x} + C_2e^{-2x} + \frac{7}{13x - 7} xe^{-2x} $$

Use initial values to determine constants

Now, we use the initial values \(y(0) = 1\) and \(y'(0) = -17\) to determine the constants \(C_1\) and \(C_2\). First, let's find y'(x) using the general solution: $$ y'(x) = 5C_1e^{5x} - 2C_2e^{-2x} + \frac{7}{13x - 7} (-2xe^{-2x}) - \frac{7}{(13x-7)^2} e^{-2x} $$ Now, use the initial values: For \(y(0) = 1\): $$ 1 = C_1e^0 + C_2e^0 + 0 $$ So, \(C_1 + C_2 = 1\). For \(y'(0) = -17\): $$ -17 = 5C_1 - 2C_2 + 0 - 0 $$ So, \(5C_1 - 2C_2 = -17\). Now we have a system of linear equations in \(C_1\) and \(C_2\). Solving it, we get \(C_1 = -4\) and \(C_2 = 5\). Therefore, the solution of the initial value problem is: $$ y(x) = -4e^{5x} + 5e^{-2x} + \frac{7}{13x - 7} xe^{-2x} $$

Plot the solution

In order to plot the solution, enter the function \(y(x) = -4e^{5x} + 5e^{-2x} + \frac{7}{13x - 7} xe^{-2x}\) into a graphing calculator or software, and visualize the graph.

Key concepts

Initial Value Problem

An Initial Value Problem, often abbreviated as IVP, is a type of differential equation paired with a specific set of conditions, known as initial conditions. These conditions allow us to find a unique solution to the differential equation. In this case, the differential equation is \[ y'' - 3y' - 10y = 7e^{-2x} \] and the initial conditions are given as \( y(0) = 1 \) and \( y'(0) = -17 \).

• The main goal of an IVP is to find a solution \( y(x) \) that satisfies both the differential equation and the initial conditions.
• By plugging in the initial conditions into the general solution of the differential equation, we determine the specific values of any arbitrary constants involved.

With the given initial conditions, these values are essential in ensuring the solution fits exactly to the problem's requirements. This process ensures that our solution is not just a family of functions, but the precise function that passes through the specific starting values at the beginning of the problem.

Complementary Function

The Complementary Function, abbreviated as \( y_c(x) \), is a critical part of solving linear differential equations. It represents the general solution to the associated homogeneous equation, which is obtained by setting the non-homogeneous part equal to zero. For our exercise, this corresponds to solving:\[ y'' - 3y' - 10y = 0 \]

• To find \( y_c(x) \), we look for solutions of the form \( e^{mx} \), and solve the resulting characteristic equation.
• The characteristic equation in the example was \( m^2 - 3m - 10 = 0 \), which is factored to \((m - 5)(m + 2) = 0\).
• The solutions \( m = 5 \) and \( m = -2 \) lead to the complementary function: \( y_c(x) = C_1e^{5x} + C_2e^{-2x} \).

The complementary function often contains arbitrary constants \( C_1 \) and \( C_2 \), which will later be determined using the initial conditions given in the problem.

Particular Solution

The Particular Solution, denoted \( y_p(x) \), addresses the non-homogeneous component of the differential equation. It adds the specific solution needed for the entire equation to accommodate the inhomogeneous part. In our example, this means finding a solution to:\[ y'' - 3y' - 10y = 7e^{-2x} \]

• We assume the form of \( y_p \) based on the non-homogeneous term (here, \( 7e^{-2x} \)), and consider a function like \( y_p(x) = Ax e^{-2x} \).
• By substituting this assumed form into the differential equation, we derive equations to solve for parameters, in this case, resolving for \( A \).
• Ensuring \( y_p(x) \) does not overlap with the complimentary function's form, we find an appropriate \( A \) and compute \( y_p(x) \).

The particular solution aligns the overall solution with the non-homogeneous part of the differential equation ensuring that all components are considered.

Homogeneous Equation

A Homogeneous Equation in differential equations is a special type where the equation equals zero, leading to simplifications. It forms the foundation for finding the Complementary Function.In our problem, the homogeneous equation was:\[ y'' - 3y' - 10y = 0 \]

• The homogeneous nature means all terms are differentials of \( y \), with no additional forcing function on the other side of the equation.
• This equation determines the dynamics and behavior of the system without external inputs, capturing its inherent properties.
• To solve it, we use characteristic equations, which potentially involve real or complex roots leading to exponential solutions as seen in the complementary function.

By solving the homogeneous part, we identify the core oscillatory modes or exponential growth/decay without external disturbances, defining the backbone of our overall solution.