Question

Find a particular solution by the method used in Example 5.3.2. Then find the general solution and, where indicated, solve the initial value problem and graph the solution. $$ y^{\prime \prime}+6 y^{\prime}+10 y=22+20 x, \quad y(0)=2, y^{\prime}(0)=-2 $$

Short answer

Question: Solve the given second-order homogeneous linear differential equation and find the solution to the given initial value problem. Differential equation: $$y^{\prime\prime} + 6y^{\prime} + 10y = 22 + 20x$$ Initial conditions: $$y(0) = 2$$ and $$y^{\prime}(0) = -2$$ Solution: 1. Find the complementary solution: $$y_c(x)=e^{-3x}(A\cos(x) + B\sin(x))$$ 2. Find a particular solution: $$y_p(x) = 2x + 1$$ 3. Find the general solution: $$y(x)=e^{-3x}(A\cos(x) + B\sin(x))+2x+1$$ 4. Solve the initial value problem: $$y(x)=e^{-3x}(\cos(x) - 3\sin(x))+2x+1$$ 5. Graph the solution using suitable software. The solution to the initial value problem is $$y(x)=e^{-3x}(\cos(x) - 3\sin(x))+2x+1$$.

Step-by-step solution

Find the complementary solution

To find the complementary solution of the homogeneous equation, we first need to find the characteristic equation associated with the given differential equation: $$ r^2+6r+10 = 0 $$ Next, we will solve this quadratic equation for r. In this case, we can use the quadratic formula: $$ r=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ Plug in the values a = 1, b = 6, and c = 10: $$ r=\frac{-6\pm\sqrt{6^2-4(1)(10)}}{2} = -3 \pm i $$ Since we have complex roots, our complementary solution will be in the form: $$ y_c(x)=e^{-3x}(A\cos(x) + B\sin(x)) $$

Finding a particular solution

To find a particular solution, we will try a solution of the form $$y_p(x) = Cx + D$$, where C and D are constants to be determined. Now, calculate the first and second derivatives of \(y_p(x)\): $$ y_p^{\prime}(x) = C\\ y_p^{\prime \prime}(x) = 0 $$ Plug these values into the original differential equation: $$ 0+6C+10(Cx+D)=22+20x $$ Comparing the coefficients on both sides, we get the following system of equations: $$ 10C = 20, \quad 6C+10D = 22 $$ Solving this system of equations, we get C = 2 and D = 1. Therefore, our particular solution is: $$ y_p(x) = 2x+1 $$

Find the general solution to the differential equation

Combine the complementary and particular solutions to get the general solution: $$ y(x)=y_c(x)+y_p(x)=e^{-3x}(A\cos(x) + B\sin(x))+2x+1 $$

Solve the initial value problem

We are given the initial conditions \(y(0)=2\) and \(y^{\prime}(0)=-2\). Plug in these values into the general solution and its derivative: $$ 2=e^{-3(0)}(A\cos(0) + B\sin(0))+2(0)+1 $$ This gives us A = 1. Now, calculate the first derivative of the general solution: $$ y^{\prime}(x)=-3e^{-3x}(A\cos(x) + B\sin(x))+e^{-3x}(-A\sin(x) + B\cos(x))+2 $$ Plugging in the initial condition \(y^{\prime}(0)=-2\), we get: $$ -2=-3(A)+B +2 $$ Knowing that A=1, we get B = -3. Thus, the solution to the initial value problem is: $$ y(x)=e^{-3x}(\cos(x) - 3\sin(x))+2x+1 $$

Graph the solution

To graph the solution, use software like Desmos, Matlab, or Python's Matplotlib library, to plot the function $$ y(x)=e^{-3x}(\cos(x) - 3\sin(x))+2x+1 $$ Choose a suitable domain for x, such as from -3 to 3, and ensure that the initial conditions at x=0 are met.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific initial conditions. These conditions help us to find a particular and unique solution that fits the scenario. In this exercise, the differential equation is given as \( y^{\prime \prime} + 6 y^{\prime} + 10 y = 22 + 20x \) with initial conditions \( y(0) = 2 \) and \( y^{\prime}(0) = -2 \). The initial conditions specify the value of the solution and its derivative at \( x = 0 \). Thus, they are crucial as they allow us to determine the constants involved in the general solution.

Complementary Solution

The complementary solution is derived from the homogeneous version of the given differential equation. Essentially, this means we set the non-homogeneous part of the equation (\( 22 + 20x \) in this case) to zero. For our example, the homogeneous equation is \( y^{\prime \prime} + 6 y^{\prime} + 10 y = 0 \). The goal is to solve this equation separately to grasp how the system behaves without external forces or inputs.
The solution involves solving a characteristic equation which results from substituting \( y = e^{rx} \) in the homogeneous equation. The resulting characteristic equation is \( r^2 + 6r + 10 = 0 \). Solving this gives complex roots \( -3 \pm i \). Therefore, the complementary solution takes the form of \( y_c = e^{-3x}(A \cos(x) + B \sin(x)) \). We use this to build the general solution.

Characteristic Equation

The characteristic equation arises from analyzing the homogeneous differential equation. It is a polynomial equation whose degree matches the order of the differential equation. In the given exercise, because we have a second-order differential equation \( y^{\prime \prime} + 6 y^{\prime} + 10 y = 0 \), its characteristic equation is a quadratic: \( r^2 + 6r + 10 = 0 \).
To find the roots of this quadratic equation, we employ the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = 10 \). This results in roots \( r = -3 \pm i \). When the roots are complex, as in this problem, the complementary solution involves exponential terms and trigonometric functions, reflecting oscillatory behavior in the system.

Particular Solution

The particular solution addresses the specific input or forcing function of the differential equation, which is the non-homogeneous part, \( 22 + 20x \). By finding a particular solution, we aim to capture the equation's response to this input.
We assume a possible form for \( y_p \), such as \( y_p = Cx + D \), based on the structure of the non-homogeneous terms. By substituting the assumed form and its derivatives back into the differential equation, we equate coefficients, yielding equations such as \( 10C = 20 \) and \( 6C + 10D = 22 \). Solving this gives \( C = 2 \) and \( D = 1 \), thus resulting in \( y_p = 2x + 1 \).
This particular solution is then combined with the complementary solution to form the general solution, from which we derive the unique solution satisfying the initial conditions.