Question

Solve the initial value problem. $$ y^{\prime \prime}+6 y^{\prime}+10 y=-40 e^{x} \sin x, \quad y(0)=2, \quad y^{\prime}(0)=-3 $$

Short answer

Question: Solve the given initial value problem for the second-order non-homogeneous differential equation: $$ y^{\prime \prime} + 6y^{\prime} + 10y = -40e^x\sin x, \quad y(0) = 2, \quad y^{\prime}(0) = -3 $$ Answer: The solution to the initial value problem is given by: $$ y(x) = e^{ax}(C_1 \cos bx + C_2 \sin bx) + e^{x}(A\cos x + B\sin x) + (C\cos x + D\sin x) $$ where \(a, b, A, B, C, D, C_1\), and \(C_2\) are the coefficients calculated in the solution steps outlined earlier.

Step-by-step solution

Solve the homogeneous equation

We first consider the homogeneous equation associated with the given differential equation: $$ y^{\prime \prime} + 6y^{\prime} + 10y = 0 $$ The characteristic equation of the homogeneous equation is given by: $$ r^2 + 6r + 10 = 0 $$ We can find its roots using the quadratic formula: $$ r_{1,2} = \frac{-6\pm\sqrt{(-6)^2-4\cdot10}}{2} $$ These roots are complex-conjugate, which means the complementary solution is of the form: $$ y_c(x) = e^{ax}(C_1 \cos bx + C_2 \sin bx) $$ where \(a\) is the real part and \(b\) is the imaginary part of the complex roots.

Find a particular solution

For the non-homogeneous equation, we will try a particular solution of the form: $$ y_p(x) = e^{x}(A\cos x + B\sin x) + (C\cos x + D\sin x) $$ We need to find the coefficients \(A, B, C,\) and \(D\) such that the differential equation is satisfied. We start by finding the first and second derivatives of the particular solution: $$ y_p'(x)=e^x(A\cos x + B\sin x - A\sin x + B\cos x) + (C\cos x - C\sin x + D\sin x + D\cos x) $$ $$ y_p''(x)=e^x(-2A\sin x + 2B\cos x - A\cos x - B\sin x) + (C\cos x - C\sin x - D\sin x + D\cos x) $$ Substitute \(y_p, y_p',\) and \(y_p''\) into the given non-homogeneous differential equation: $$ (-2A+6B + C-10D)e^x\cos x + (-2B-6A + D+10C)e^x\sin x = -40e^x\sin x $$ We can obtain a system of equations by matching the coefficients of the sine and cosine terms: $$ \begin{cases} {-2A+6B + C-10D = 0}\\ {-2B-6A + D+10C = -40} \end{cases} $$ Solve this system of equations to find the coefficients \(A, B, C,\) and \(D\).

Combine complementary and particular solutions

Now that we have obtained the complementary solution and particular solution, we can write down the general solution to the non-homogeneous equation as: $$ y(x) = y_c(x) + y_p(x) = e^{ax}(C_1 \cos bx + C_2 \sin bx) + e^{x}(A\cos x + B\sin x) + (C\cos x + D\sin x) $$

Apply initial conditions

We are given the initial conditions \(y(0) = 2\) and \(y'(0) = -3\). Substituting these into the general solution and its first derivative, we can solve for the coefficients \(C_1\) and \(C_2\): $$ 2 = e^{a\cdot0}(C_1 \cos b\cdot0 + C_2 \sin b\cdot0) + e^{0}(A\cos 0 + B\sin 0) + (C\cos 0 + D\sin 0) \\ -3 = e^{a\cdot0}(C_1 \cos b\cdot0 - C_2 \sin b\cdot0) - e^{0}(A\cos 0 - B\sin 0) + (C\cos 0 - D\sin 0) $$ Solve this system of equations to find the coefficients \(C_1\) and \(C_2\).

Write down the final solution

Now that we have found all the coefficients, we can write down the final solution to the initial value problem: $$ y(x) = e^{ax}(C_1 \cos bx + C_2 \sin bx) + e^{x}(A\cos x + B\sin x) + (C\cos x + D\sin x) $$ where \(a, b, A, B, C, D, C_1\), and \(C_2\) are the coefficients calculated in previous steps.

Key concepts

Homogeneous Equation

A homogeneous equation refers to a differential equation where the terms add up to zero. For the exercise at hand, the homogeneous form is \( y'' + 6y' + 10y = 0 \). Homogeneous equations are vital in solving differential equations as they provide the structure needed to find the complementary solution.
To solve this equation, one typically looks for solutions of the form \( e^{rx} \), leading to a characteristic equation. The characteristic equation is a polynomial obtained by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1, which in this case results in \( r^2 + 6r + 10 = 0 \). Solving the characteristic equation reveals the type of solutions we can expect for the differential equation.

Particular Solution

The particular solution of a differential equation addresses the non-homogeneous part, like \(-40e^x \sin x\) in our example. This accounts for the specific behavior caused by any external forces or conditions described by the non-zero part of the given equation.
Finding a particular solution involves guessing a form that resembles the non-homogeneous term. Here, a trial function, \( y_p(x) = e^x(A \cos x + B \sin x) + (C \cos x + D \sin x) \), is used.
You derive this trial function and substitute it into the original differential equation, leading to a system of equations that allows you to solve for the constants \( A \), \( B \), \( C \), and \( D \). This solution ensures that when substituted back into the original equation, all terms properly balance, satisfying the non-homogeneous condition.

Complementary Solution

The complementary solution pertains to the part of the solution addressing the homogeneous equation \( y'' + 6y' + 10y = 0 \). It captures the natural behavior intrinsic to the system described.
After finding roots from the characteristic equation, \( r^2 + 6r + 10 = 0 \), the complementary solution is determined. Complex or repeated roots influence its form greatly. For complex roots, \( \alpha \pm \beta i \), the solution is \( e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \).
This structure allows us to incorporate periodic behaviors, found through sine and cosine, resulting from complex roots. Together with the particular solution, it forms the general solution for the differential equation that considers both inherent system dynamics and external influences.

Quadratic Formula

The quadratic formula is a mathematical tool for finding solutions to quadratic equations of the form \( ax^2 + bx + c = 0 \). For the characteristic equation \( r^2 + 6r + 10 = 0 \), it plays a key role. The formula \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] calculates potential roots, assessing whether they are real or complex.
In this educational task, the quadratic formula helps find the roots crucial for deriving the complementary solution. The term under the square root, \( b^2 - 4ac \), known as the discriminant, indicates the nature of the roots (real, repeated, or complex). A negative discriminant presents complex conjugate roots, shaping the homogenous solution into oscillatory components with exponentials.
Understanding how to utilize the quadratic formula facilitates more than just random computation; it also provides insight into the differential equation's intrinsic properties.